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Math & DP & Data structure & Graph & Geometry & Game

表演者: magicat ?? nannan

- Math & DP & Data structure & Graph & Geometry & Game
Head
Math
DP
Data structure
Graph
計算幾何
大數
雜
- 內建函數
過去犯錯

Head

快讀1

ios::sync_with_stdio(false);   cin.tie(nullptr), cout.tie(nullptr);

快讀2

int rd() {
    char act = 0;
    int f = 1, x = 0;
    while (act = getchar(), act < '0' && act != '-')
    ;
    if (act == '-') f = -1, act = getchar();
    x = act - '0';
    while (act = getchar(), act >= '0') x = x * 10 + act - '0';
    return x * f;
}

快讀3

char nc() {
    static char buf[1000000], *p = buf, *q = buf;
    return p == q && (q = (p = buf) + fread(buf, 1, 1000000, stdin), p == q)
    ? EOF
    : *p++;
}

int rd() {
    int res = 0;
    char c = nc();
    while (!isdigit(c)) c = nc();
    while (isdigit(c)) res = res * 10 + c - '0', c = nc();
    return res;
}

模板1

// AC one more times

#include <bits/stdc++.h>

using namespace std;

typedef long long ll;

const int mod = 1e9 + 7;
const int N = 2e5 + 10;


void solve()
{       


    return;
}


  
int main()
{
    ios::sync_with_stdio(false);   cin.tie(nullptr), cout.tie(nullptr);
    
    int TC = 1;
    
    cin >> TC;    
    for(int tc = 1; tc <= TC; tc++)
    {
        //cout << "Case #" << tc << ": ";         
        solve();
    }


    return 0;
}

模板2

// AC one more times
// nndbk
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int mod = 1e9 + 7;
const int N = 2e5 + 10;

int main()
{
    ios::sync_with_stdio(false);   cin.tie(nullptr), cout.tie(nullptr);
  


    return 0;
}

模板3

// AC one more times



////////////////////////////////////////INCLUDE//////////////////////////////////////////

#include <bits/stdc++.h>

using namespace std;
/////////////////////////////////////////DEFINE//////////////////////////////////////////

#define fi first
#define se second
#define pb push_back
#define endl '\n'
#define all(x) (x).begin(), (x).end()
#define inf64 0x3f3f3f3f3f3f3f3f

typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> pii;
typedef pair<long long,long long> pll;

////////////////////////////////////////CONST////////////////////////////////////////////

const int inf = 0x3f3f3f3f;
const int maxn = 2e6 + 6;

///////////////////////////////////////FUNCTION//////////////////////////////////////////

int gcd(int a, int b) { return b == 0 ? a : gcd(b, a % b); }
int lcm(int a, int b) { return a / gcd(a, b) * b; }
inline __int128 read(){__int128 x=0,f=1;char ch=getchar();while(ch<'0'||ch>'9'){if(ch=='-'){f=-1;}ch=getchar();}while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}return x*f;}
inline void write(__int128 x){if (x < 0){putchar('-');x = -x;}if (x > 9) write(x / 10);putchar(x % 10 + '0');}
ll qmi(ll a,ll b,ll p){   ll ans=1%p;while(b){    if(b&1) {ans=ans*a%p;}  a=a*a%p,b>>=1;    }    return ans;  }
int dx[]={0, 0, -1, 1};
int dy[]={-1, 1, 0, 0};


const double eps = 1e-8;
const int mod = 1e9 + 7;
const int N = 2e5 + 10;



void solve()
{   

    return;
}

  
int main()
{
    std::ios::sync_with_stdio(false);   cin.tie(nullptr), cout.tie(nullptr);
  
    int TC = 1;
    
    cin >> TC;    
    for(int tc = 1; tc <= TC; tc++)
    {
        //cout << "Case #" << tc << ": ";         
        solve();
    }


    return 0;
}

Math

快速冪

ll qmi(ll a, ll b, ll mod)
{
    ll ans = 1 % mod;
    while(b)
    {
        if(b & 1) ans = ans * a % mod;
        a = a * a % mod;
        b >>= 1;
    }
    return ans;
}

快速乘

//乘法分配率快速乘，時間復雜度O(1)
ll mul(ll a,ll b, ll P){
    ll L = a * (b >> 25ll) % P * (1ll << 25) % P;
   	ll R = a * (b & ((1ll << 25) - 1)) % P;
    return (L + R) % P;
}
/*
我們要計算a*b mod p,設b = L+R
那么原式變為a*(L+R)mod p = ((a*L)mod p+(a*R)mod p)mod p
我們定L為b的二進制前x位，R為b的后(64-x)位
這樣得到的代碼(以上代碼x = 25),復雜度近似為O(1).
*/

除法取整

ll floordiv(ll a, ll b)
{
    //因為當其是負數時候，c++默認往0取整，所以下取整我們要自己寫
    if(a % b == 0)  return a / b;
    else if(a > 0)  return a / b;
    else return a / b - 1;
}

ll ceildiv(ll a, ll b)
{
    if(a % b == 0)  return a / b;
    else if(a > 0)  return a / b + 1;
    else return a / b;
}

龜速乘

ll a,b,p;
ll qmimul(ll a, ll b, ll p)
{
    ll ans=0;
    a %= p;
    for(; b; b >>= 1)
    {
        if(b & 1)
            ans += a, ans %= p;
        a += a, a %= p;
    }
    return ans;
}
void solve()
{
    cin>>a>>b>>p;    cout<<qmimul(a,b,p)<<endl;
}

數論分塊

void solve()
{
    ll n, ans = 0;  cin>>n;
    for(ll l = 1; l <= n; l++)
    {
        ll d = n / l, r = n / d;
        // ans + ...
        l = r;
    }
    return;
}

枚舉子集，超集

void solve()
{
    int n;  cin>>n;
    // 枚舉子集
    for(int i = 1; i < (1 << n); i++)
        for(int j = i; j; j = (j - 1) & i)
        {

        }
    // 枚舉超集
    for(int i = 0; i < (1 << n); i++)
        for(int j = i; ; j = (j + 1) | i)
        {

            if(j == (1 << n) - 1)
                break;
        }
}

矩陣乘法

const int mod = 1e9+7;
const int N = 200;
int n, m;  
long long a[N + 1][N + 1], f[N + 1];
void aa()
{
    long long w[N + 1][N + 1];
    memset(w, 0, sizeof w);
    for(int i = 1; i <= N; i++)
        for(int k = 1; k <= N; k++)
            if(a[i][k])
                for(int j = 1; j <= N; j++)
                    if(a[k][j])
                        w[i][j] += a[i][k] * a[k][j], w[i][j] %= mod;
    memcpy(a, w, sizeof a);
}

void fa()
{
    long long w[N + 1];
    memset(w, 0, sizeof w);
    for(int i = 1; i <= N; i++)
        for(int j = 1; j <= N; j++)
            w[i] += f[j] * a[j][i], w[i] %= mod;
    memcpy(f, w, sizeof f);
}

void matrixpow(long long k)
{
    for(; k; k /= 2)
    {
        if(k & 1)
            fa();
        aa();
    }
}

void solve()
{
    cin>>n>>m;
    ll k;  cin>>k;
    // construct Martix
    
    
    matrixpow(k);
    cout<<f[ ]<<endl;
    return;
}

矩陣運算及求行列式

#include <bits/stdc++.h>
 
using namespace std;
 
typedef long long ll;
 
struct Matrix {
    ll N;
    vector<vector<ll>> A;
    Matrix() { N = 0;}
    Matrix(int n) {
        N = n;
        A.resize(N + 1);
        for (int i = 0; i <= N; i ++)
            A[i].resize(N + 1, 0);
    }
    void init(vector<vector<ll>> a, int n) {
        A = a;
        N = n;
    }
 
    Matrix operator*(const Matrix &b) const {
        Matrix res(b.N);
 
        for (int i = 1; i <= b.N; ++i)
            for (int j = 1; j <= b.N; ++j)
                for (int k = 1; k <= b.N; ++k)
                    res.A[i][j] = (res.A[i][j] + A[i][k] * b.A[k][j]);
        return res;
    }
    Matrix qpow(ll k) {
        Matrix res(N);
 
        //斐波那契數列初始化
        //res.A[1][1] = res.A[1][2] = 1;
        //A[1][1] = A[1][2] = A[2][1] = 1;
 
        //單位矩陣
        for (int i = 0; i <= N; i ++)
            res.A[i][i] = 1;
 
        while (k) {
            if (k & 1) res = res * (*this);
            (*this) = (*this) * (*this);
            k >>= 1;
        }
        return res;
    }
    //求行列式
    ll det() {
        return DET(A, N);
    }
 
    ll DET(vector<vector<ll>> arr1, int n) {
        ll sum = 0;
        //i是第一行的列指標，M是余子式的值，sum是行列式的計算值
        if (n == 1)//一階行列式直接得出結果
            return arr1[0][0];
        else if (n > 1) {
            for (int i = 0; i < n; i++) {
                //按照第一行展開
                ll M = Minor(arr1, i, n);
                sum += pow(-1, i + 2) * arr1[0][i] * M;
            }
        }
        return sum;
    }
    ll Minor(vector<vector<ll>> arr1, int i, int n) {
        vector<vector<ll>>arr2(n-1,vector<ll>(n-1));
        //以下為構造余子式的過程。
        for (int j = 0; j < n - 1; j++) {
            for (int k = 0; k < n - 1; k++) {
                if (k < i)
                    arr2[j][k] = arr1[j + 1][k];
                else if (k >= i)
                    arr2[j][k] = arr1[j + 1][k + 1];
            }
        }
 
        return DET(arr2, n - 1);
        //構造完后，余子式是一個新的行列式，返回DET函數進行計算。
    }
};
 
int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
 
    int n, m;
    cin >> n >> m;
 
    vector<vector<ll>>a(max(n,m),vector<ll>(max(n,m),0));


    for (int i = 0; i < n; i ++)
        for (int j = 0; j < m; j ++)
            cin >> a[i][j];
 
    ll ans = 1e9;
 
    Matrix Ma;
    for (int i = 1; i <= max(n, m); i ++) {
        vector<vector<ll>>ok(i,vector<ll>(i,0));

        for (int j = 0; j < i; j ++)
            for (int k = 0; k < i; k ++)
                ok[j][k] = a[j][k];
        Ma.init(ok, i);
        ans = min(ans, Ma.det());
    }
 
    cout << ans << '\n';
 
    return 0;
}

線性篩

int tot, p[N], pr[N];
void prime(int n)
{
    for (int i = 2; i <= n; i++)
    {
        if (!p[i])    p[i] = i, pr[++tot] = i; 
        for (int j = 1; j <= tot && pr[j] * i <= n; j++) {
            p[pr[j] * i] = pr[j];
            if(p[i] == pr[j])   break;
        }
    }
}

const int N =1000010;
bool is_pri[N];
int pri[N];
int cnt;

void init(int n)
{
	memset(is_pri, true, sizeof(is_pri));
	is_pri[1] = false;
	cnt = 0;
	for (int i = 2; i <= n; i++)
	{
		if (is_pri[i])
			pri[++cnt] = i;
		for (int j = 1; j <= cnt && i * pri[j] <= n; j++)
		{
			is_pri[i * pri[j]] = false;
			if (i % pri[j] == 0)break;            
		}
	}
}

埃式篩

int tot, p[N], pr[N];
void prime(int n)
{
    for(int i = 2; i <= n; i++)
    {
        if(p[i])    continue;
        pr[++tot] = i;  
        for(int j = i; j <= n / i; j++)     p[i * j] = 1;
    }
}

const int N = 1e5;		
bool isPrime[N + 5];	
int primes[N + 5];		
int cnt = 0;	

void aiPrime()
{
	memset(isPrime,true,sizeof(isPrime));		
 
	for (int i = 2;i <= N / i;i++)//注意這里 i <= N / i就可以
	{
		if (isPrime[i])		//如果當前數是質數，那么將它的倍數都標記為 false
		{			
			for (int j = i * i; j <= N; j += i)
			{
				isPrime[j] = false;
			}
		}
	}
 
	for (int i = 2; i <= N; i++)		//將質數放入primes數組
	{
		if(isPrime[i])
			primes[cnt++] = i;
	}
}

區間篩

int tot, p[N], pr[N];
bool vis[N];
void prime(int n)
{
    for (int i = 2; i <= n; i++)
    {
        if (!p[i])    p[i] = i, pr[++tot] = i; 
        for (int j = 1; j <= tot && pr[j] * i <= n; j++) {
            p[pr[j] * i] = pr[j];
            if(p[i] == pr[j])   break;
        }
    }
}

void solve()
{   
    int l, r;   cin>>l>>r;
    prime(sqrt((1ll << 31)) + 10);
    for(int i = 1; i <= tot; i++)
    {
        int p = pr[i];
        for(int x = r / p * p; x >= l; x -= p)
            if(x != p)
                vis[x - l] = true;
    }

    int res = 0;
    for(ll x = l; x <= r; x++)
        if(!vis[x - l])
            res++;
    if(l == 1)  res--;
    cout<<res<<'\n';
    return;
}

質因數分解

ll a[N],cnt;
void primer(ll x)
{
    for (ll i = 2; i <= x / i; i++)
    {
        if (x % i == 0)
        {
            int s = 0;
            a[++cnt] = i;
            while (x % i == 0)
            {
                x = x / i;
                s++;
            }
        }
    }
    if (x > 1) a[++cnt] = x;
}

歐拉函數

ll phi(ll n)
{
	ll phin = n;
	for(ll i = 2; i <= n / i; i++)
	{
		if(n % i == 0)
		{
			phin = phin / i * (i - 1);
			while(n % i == 0)
				n /= i;
		}
	}
	if(n != 1)
		phin = phin / n * (n - 1);
	return phin;
}

約數

最大公約數

int gcd(int a, int b)
{
    return b ? gcd(b, a % b) : a;
}

求約數

vector<int> get_yueshu(int x)
{
    vector<int> a;
    for (int i = 1; i <= x / i; i++)
    {
        if (x % i == 0)
        {
            a.push_back(i);
            if (i != x / i) 
                a.push_back(x / i);
        }
    }
    sort(a.begin(), a.end());
    return a;
}

擴展歐幾里得1

$ax - by = \gcd(a, b)$ 的最小非負整數解 $(x,y)$

ll exgcd(ll a, ll b, ll &x, ll &y)
{
    if(b == 0)
    {
        x = 1; y = 0;
        return a;
    }
    ll xx,yy;
    ll d = exgcd(b, a % b, xx, yy);
    x = yy;
    y = xx - (a / b) * yy;
    return d;
}

int main()
{
    int _;cin>>_;
    while(_--)
    {
        int a, b, x, y;
        cin>>a>>b;
        int d = exgcd(a, b, x, y);
        y = -y;
        while(x < 0 || y < 0)   x += b / d, y += a / d;
        while(x >= b / d && y >= a / d) x -= b / d, y -= a / d;
        cout<<x<<" "<<y<<endl;
    }
}

擴展歐幾里得2

$ax + by = d$ 的 $x$ 最小的解的非負整數解 $(x,y)$

ll exgcd(ll a, ll b, ll &x, ll &y)
{
    if(b == 0)
    {
        x = 1; y = 0;
        return a;
    }
    ll xx,yy;
    ll d = exgcd(b, a % b, xx, yy);
    x = yy;
    y = xx - (a / b) * yy;
    return d;
}

int main()
{
    int _;cin>>_;
    while(_--)
    {
        ll a, b, x, y, m;
        scanf("%lld%lld%lld", &a, &b, &m);
        ll d = exgcd(a, b, x, y);
        if(m % d != 0)
        {
            puts("-1");
            continue;
        }
        a /= d;
        b /= d;
        m /= d;
        ll xx = (ll) x * (m % b) % b;
        if(xx < 0)  xx = xx + b;
        ll yy = (ll)(m - a * xx) / b;
        if(yy < 0)
            puts("-1");
        else
            printf("%lld %lld\n", xx, yy);
    }
}

線性同余方程

ll modequ(ll a, ll b, ll m) {
    ll x, y;
    ll d = exgcd(a, m, x, y);
    if (b % d != 0) return -1;
    m /= d; a /= d; b /= d;
    x = x * b % m;
    if (x < 0) x += m;
    return x;
}

【模板】二元一次不定方程 (exgcd)

求正整數解個數和 $xmin,xmax,ymin,ymax$ 的，因為要求是正的，注意判等于 $0$ 的情況

給定不定方程

\[ax+by=c \]

若該方程無整數解，輸出 $-1$。
若該方程有整數解，且有正整數解，則輸出其正整數解的數量，所有正整數解中 $x$ 的最小值，所有正整數解中 $y$ 的最小值，所有正整數解中 $x$ 的最大值，以及所有正整數解中 $y$ 的最大值。
若方程有整數解，但沒有正整數解，你需要輸出所有整數解中 $x$ 的最小正整數值， $y$ 的最小正整數值。

正整數解即為 $x, y$ 均為正整數的解，$\boldsymbol{0}$ 不是正整數。
整數解即為 $x,y$ 均為整數的解。
$x$ 的最小正整數值即所有 $x$ 為正整數的整數解中 $x$ 的最小值，$y$ 同理。

$1 \le a, b, c \le {10}^9$。

sample

4 6 2 39 8
2 1
-1
1600 1 18 3199 30399
34 3
-1
2 12 7 50 56

typedef long long ll;
ll exgcd(ll a, ll b, ll &x, ll &y)
{
    if(b==0)
    {
        x = 1, y = 0;
        return a;
    }
    ll d = exgcd(b, a % b, y, x);
    y -= (a / b) * x;
    return d;
}
int main()
{
    ll a, b, c, x, y;
    cin>>a>>b>>c;
    ll d = exgcd(a, b, x, y);
    if(c % d)
        cout<<-1<<"\n";
    else
    {
        a /= d, b /= d, c /= d;
        x *= c,y *= c;//特解
        x = (x % b + b) % b;
        x = x == 0 ? b : x;
        y = (c - a * x) / b;
        if(y > 0)
        {
            ll xmin, xmax, ymin, ymax;
            xmin = x,ymax = y;
            y %= a;
            y = y == 0 ? a : y;
            x = (c - b * y) / a;
            xmax = x, ymin = y;
            ll cnt = (xmax - xmin) / b + 1;//每隔b一個整數解
            cout<<cnt<<" "<<xmin<<" "<<ymin<<" "<<xmax<<" "<<ymax<<"\n";
        }
        else//有整數解，但無正整數解
        {
            ll xmin, ymin;
            xmin = x;
            y = y % a + a;
            ymin = y;
            cout<<xmin<<" "<<ymin<<"\n";
        }
    }
    return 0;
}

中國剩余定理

dls

ll exgcd(ll a, ll b, ll &x, ll &y)
{
    if(b == 0)
    {
        x = 1; y = 0;
        return a;
    }
    ll xx,yy;
    ll d = exgcd(b, a % b, xx, yy);
    x = yy;
    y = xx - (a / b) * yy;
    return d;
}

void merge(ll &a, ll &b, ll c, ll d) { // d <= 10^9
    // bt=c-a(mod d)
    if (a == -1 && b == -1) return;
    ll x, y;
    ll g = exgcd(b, d, x, y);
    //bx=g(mod d)
    if ((c - a) % g != 0) {
        a = b = -1;
    }
    d /= g; // d'
    ll t0 = ((c - a) / g) % d * x % d;
    if (t0 < 0) t0 += d;
    // t = t0 (mod d')
    a = b * t0 + a;
    b = b * d;
}

ex中國剩余定理

typedef long long ll;

const int mod = 19260817;
const int N = 1e8 + 10;

__int128 exgcd(__int128 a, __int128 b, __int128 &x, __int128 &y)
{
    if(b == 0)
    {
        x = 1; y = 0;
        return a;
    }
    __int128 xx,yy;
    __int128 d = exgcd(b, a % b, xx, yy);
    x = yy;
    y = xx - (a / b) * yy;
    return d;
}

void merge(__int128 &a, __int128 &b, __int128 c, __int128 d) { // d <= 10^9
    // bt=c-a(mod d)
    if (a == -1 && b == -1) return;
    __int128 x, y;
    __int128 g = exgcd(b, d, x, y);
    //bx=g(mod d)
    if ((c - a) % g != 0) {
        a = b = -1;
    }
    d /= g; // d'
    __int128 t0 = ((c - a) / g) % d * x % d;
    if (t0 < 0) t0 += d;
    // t = t0 (mod d')
    a = b * t0 + a;
    b = b * d;
}


array<ll, 2> ar[N];
void solve()
{   
    int n;  cin>>n;
    for(int i = 1; i <= n; i++)
        cin>>ar[i][1]>>ar[i][0];
    __int128 a = 0, b = 1;
    for(int i = 1; i <= n; i++)
        merge(a, b, (__int128)ar[i][0], (__int128)ar[i][1]);
    cout<<(ll)a<<'\n';



    return;
}

逆元

線性

ll n, p, inv[N];
void solve()
{
    cin>>p>>n;
    inv[1] = 1;
    for(int i = 2; i <= n; i++)
        inv[i] = (p - p / i) * inv[p % i] % p;
    return;
}

快速冪求逆元

qmi(a, mod - 2, mod)

階 $a ^ x \equiv 1 (\bmod p)$

int p, phip;
vector<int> factor;
ll qmi(ll a, ll b, ll p)
{
	ll res = 1;
	for(; b; b >>= 1)
	{
		if(b & 1) res = res * a % p;
		a = a * a % p;	
	}
	return res;
}

int phi(int p)
{
	int res = p;
	for(int i = 2; i <= p / i; i++)
		if(p % i == 0)
			res = res / i * (i - 1);
	if(p != 1)
		res	= res / p * (p - 1);
	return res;
}

void solve()
{       
	int a;	cin>>a;
	int res = phip;
	for(auto &it : factor)
		if(qmi(a, res / it, p) == 1)
			res /= it;
	cout<<res<<'\n';
    return;
}

void init()
{
	phip = phi(p);
	int q = phip;
	for(int i = 2; i <= q / i; i++)
		if(q % i == 0)	while(q % i == 0)	factor.push_back(i), q /= i;
	if(q != 1)	factor.push_back(q);
}

原根 $a ^ x \equiv 1 (\bmod m)$ ，當最小的 $x$ 是 $\varphi(m)$ 的時候 $a$ 是 $m$ 的原根

原根個數: 若 $m$ 有原根， $m$ 有原根 $\varphi (\varphi (m))$個

最小原根的數量級: 若 $m$ 有原根，最小原根不多于 $m ^ {0.25}$ 級別

ll qmi(ll a, ll b, ll p)
{
	ll res = 1;
	for(; b; b >>= 1)
	{
		if(b & 1) res = res * a % p;
		a = a * a % p;	
	}
	return res;
}

int p, phip;
vector<int> factor;

void init()
{
	int q = p - 1;
	for(int i = 2; i <= q / i; i++)
		if(q % i == 0)	while(q % i == 0)	factor.push_back(i), q /= i;
	if(q != 1)	factor.push_back(q);
}

void solve()
{       
	factor.clear();
	cin>>p;
	phip = p - 1;
	init();
	for(int i = 1; i < p; i++)
	{
		bool ok = true;
		for(auto &it : factor)
		{
			if(qmi(i, phip / it, p) == 1)
			{
				ok = false;
				break;
			}
		}
		if(ok)
		{
			cout<<i<<'\n';
			return;
		}
	}
    return;
}

指數方程1 $a ^ x \equiv b (\bmod m)$ $m$ 為質數

ll qmi(ll a, ll b, ll p)
{
	ll res = 1;
	for(; b; b >>= 1)
	{
		if(b & 1) res = res * a % p;
		a = a * a % p;	
	}
	return res;
}

void solve()
{       
	int a, b, m;	cin>>a>>b>>m;
	int res = m + 1, T = sqrt(m) + 2;
	ll d = 1, v = qmi(a, T, m);
	unordered_map<int, int> mp;
	for(int q = 1; q <= T; q++)
	{
		d = v * d % m;
		if(mp.count(d))	continue;
		mp[d] = q * T;
	}
	d = b;
	for(int r = 1; r <= T; r++)
	{
		d = a * d % m;
		if(mp.count(d))
			res = min(res, mp[d] - r);
	}
	if(res == m + 1)
		res = -1;
	cout<<res<<'\n';
	return;
}

指數方程2 $a ^ x \equiv b (\bmod m)$ $m$ 不為質數

ll ksm(ll a, ll b, ll mod)
{
    ll ans = 1, base = a;
    while(b)
    {
        if(b & 1)   ans = ans * base % mod;
        base = base * base % mod;
        b >>= 1;
    }
    return ans;
}

ll exgcd(ll a, ll b, ll &x, ll &y)
{
    if(b == 0)
    {
        x = 1, y = 0;
        return a;
    }
    ll d = exgcd(b, a % b, y, x);
    y -= (a / b) * x;
    return d;
}

//求a*x = b(mod m)的解
ll modequ(ll a, ll b, ll m)
{
    ll x, y;
    ll d = exgcd(a, m, x, y);
    if(b % d)   return -1;
    m /= d, a /= d, b /= d;
    x = x * b % m;
    if(x < 0)   x += m;
    return x;
}

int MAGIC = 30;
bool ok = false;
/*
21 48 49
0 0 0
*/
void solve()
{
    int a, b, m;    
    cin>>a>>m>>b;
    //cout<<a<<" "<<b<<"  "<<m<<'\n';
    if(a == 0 && b == 0 && m == 0)
    {
        ok = true;
        return;
    }
    b %= m;
    if(b == 1 || m == 1)
    {
        cout<<0<<'\n';
        return;
    }
    ll v = 1;
    for(int i = 0;i < MAGIC; i++)
    {
        if(v == b)
        {
            cout<<i<<'\n';
            return;
        }
        v = v * a % m;
    }
    //v * a^{x-30} = b(mod m);
    ll g = __gcd(v, 1ll * m);
    if(b % g)
    {
        cout<<"No Solution"<<'\n';
        return;
    }
    b /= g, m /= g;
    a %= m;
    //v/g * ? = b/g(mod m)
    b = modequ(v / g, b, m);
    //BSGS
    int T = sqrt(m) + 2;//小心精度誤差，我們+2
    v = ksm(a, T, m);
    ll d = 1;
    unordered_map<int, int>ha;
    for(int q = 1; q <= T; q++)
    {
        d = d * v % m;
        //因為我們要求較小的解，如果我們有兩個相同的d，去小的那個
        if(!ha.count(d))    ha[d] = q * T;
    }
    int ans = m + 1;
    d = b;
    for(int r = 1; r <= T; r++)
    {
        d = d * a % m;
        if(ha.count(d)) ans = min(ans, ha[d] - r);
    }
    if(ans >= m)    cout<<"No Solution"<<'\n';
    else    cout<<ans + MAGIC<<'\n';

}

排列組合

加法 & 乘法原理

加法原理

完成一個工程可以有$n$ 類辦法，$a_i(1\leq i\leq n)$ $a_i(1 \le i \le n)$ 代表第 $i$ 類方法的數目。那么完成這件事共有 $S = a_1 + a_2 + \dots + a_n$ $S=a_1+a_2+\cdots +a_n$ 種不同的方法。

乘法原理

完成一個工程需要分 $n$ 個步驟，$a_i(1 \leq i \leq n)$ 代表第 $i$ 個步驟的不同方法數目。那么完成這件事共有 $S = a_1 \times a_2 \times \cdots \times a_n$ 種不同的方法。

組合數

$O(N^2)$

int n, c[N][N];
void init()
{
    for(int i = 0; i <= 2000; i++)
        for(int j = 0; j <= i; j++)
            if(j == 0)    c[i][j] = 1;
            else c[i][j] = (c[i - 1][j - 1] + c[i - 1][j]) % mod;
}
void solve()
{
    int a,b;    cin>>a>>b;
    cout<<c[a][b]<<endl;
}

O(N)

int n;
ll f[N],infact[N];
void init()
{
    f[0] = infact[0] = 1;
    for(int i = 1; i <= 100000; i++)
    {
        f[i] = i * f[i - 1] % mod;
        infact[i] = infact[i - 1] * 1ll * qmi(i, mod-2, mod) % mod;
    }
}
void solve()
{
    int a,b;    cin>>a>>b;
    cout<<(f[a] * infact[b] % mod * infact[a - b] % mod)<<endl;
}

Lucas $O(\log_pNp\log_p) p 為質數$

int C(int a, int b, int p)
{
    if (b > a) return 0;
    int res = 1;
    for (int i = 1, j = a; i <= b; i ++, j -- )
    {
        res = (ll)res * j % p;
        res = (ll)res * qmi(i, p - 2, p) % p;
    }
    return res;
}
int lucas(ll a, ll b, int p)
{
    if (a < p && b < p) return C(a, b, p);
    return (ll)C(a % p, b % p, p) * lucas(a / p, b / p, p) % p;
}

Lucas $O(\log_qNq\log_q)$

/*
組合數取模(模數p可以是合數的情況)

p是合數的情況不好處理，
像之前學的CRT，如果p是合數，我們拆成若干素數冪的形式。
比如：
n
m  mod q 

q = p1^e1*p2^e2...pk^ek
n
m mod pi^ei
再用CRT可以解決


即：我們要解決
n
m mod pi^ei  (p^e<=1e6)

CRT
        m1,m2...mn
        找到xi滿足:xi mod mi = 1
                   xi mod M/mi = 0;
        x ≡ ai(mod mi)
        x = Σxi*ai
x就是答案

Q:考慮為什么我們不能用之前的逆元的寫法
A:逆元寫法：
    n!/(m!*(n-m)!)
    a/b mod p ==> a mod p*(b mod p)^-1
    即:
    1.n!*(m!)^-1*(n-m)^-1
    2.inv[i] = (p-p/i)*inv[p%i]%p

1.因為n可能很大，直接算階乘可能算不出來
2.算逆元的話，分母不一定是可逆的,比如p=2，分母有很多2的因子,所以說它不一定是可逆的
整體思路就是把階乘都用n！= p^an*bn表示出來,其中p不整除bn(把p的因子都提出來,那這個bn相當于是多出來的)
n！= p^an*bn
= p^an*bn/(p^am*bm*p^(an-m)*(bn-m))
= p^(an-am-(an-m))*bn/(bm*(bn-m))
這里b是沒有p的因子,肯定與p的冪次是互質的,那我們可以直接求其逆元
= p^(an-am-(an-m))*bn*(bm*(bn-m))^(-1) (mod p^e)
畫個表...
n！= p^an*bn
預處理：
fac[i]：1~i之間不被p整除%p^e的所有數的乘積
fac[p^e-1]^(n/p^e)*fac[n%p^e]
有(n/p^e)個完整的組，和n%p^e多出來的零頭
遞歸下去就是：p^(n/p)*(n/p)!————>(n/p)!就是子問題
一直遞歸到<p為止
*/

#include<bits/stdc++.h>
//#pragma GCC diagnostic error "-std=c++11"
#pragma GCC optimize(2)
#pragma GCC optimize(3)
using namespace std;
 
typedef long long ll;
const int N = 101000;
int m, T, M, phipe;
pair<int, int> x[110];
ll pr[110];
ll ans[N], a[N], b[N], fac[1010000];

ll ksm(ll a, ll b, ll m)
{
    ll ans = 1, base = a;
    while(b > 0)
    {
        if(b & 1)
        {
            ans *= base;
            ans %= m;
        }
        base *= base;
        base %= m;
        b >>= 1;
    }
    return ans;
}
pair<ll, ll> calc(ll a, int p, int pe)
{
    ll cntp = 0, cnts = 0, val = 1;
    while(a)
    {
        cntp += a / p;
        cnts += a / pe;
        val = val * fac[a % pe] % pe;
        //val = val*ksm(fac[pe],a/pe)%pe*fac[a%pe]%pe
        a /= p;
    }
    //val = val*ksm(fac[pe],cnts,pe)%pe;
    val = val * ksm(fac[pe - 1], cnts % phipe, pe) % pe;
    return {val, cntp};//非素數部分和素數部分
}

ll binom(ll a, ll b, int p, int pe)
{
    /*
        n！= p^an*bn
        = p^an*bn/(p^am*bm*p^(an-m)*(bn-m))
        = p^(an-am-(an-m))*bn/(bm*(bn-m))
        = p^(an-am-(an-m))*bn*(bm*(bn-m))^(-1) (mod p^e)
    */
    auto f1 = calc(a, p, pe);
    auto f2 = calc(b, p, pe);
    auto f3 = calc(a - b, p, pe);
    ll v1 = f1.first * ksm(f2.first * f3.first % pe, phipe - 1, pe) % pe;//與p互質的部分
    ll v2 = ksm(p, f1.second - f2.second - f3.second, pe);//與p不互質的部分
    return v1 * v2 % pe;
}

int main()
{
    ios::sync_with_stdio(false);
    cin.tie(nullptr);   cout.tie(nullptr);
    cin>>m>>T;
    M = m;
    int t = 0;
    for(int i = 2; i <= m; i++)
    {
        if(m % i == 0)//對m進行分解
        {
            int p = i, pe = 1;
            while(m % i == 0)   m /= i, pe *= i;
            x[t++] = {p, pe};//x記錄分解結果
        }
    }
    /*
        CRT
        m1,m2...mn
        找到xi滿足:xi mod mi = 1
                   xi mod M/mi = 0;
        x ≡ ai(mod mi)
        x = Σxi*ai
    */
    for(int i = 0; i < t; i++)
    {
        int pe = x[i].second;
        for(int c = 0; c < M; c++)//這里M很小我們可以直接暴力for一遍找xi，就不用寫exgcd了
        {
            if(c % pe == 1 && c % (M / pe) == 0)
            {
                pr[i] = c;
                break;
            }
        }
    }
    for(int i = 0; i < T; i++)  cin>>a[i]>>b[i];//先讀進來再每個因子分開處理
    for(int i = 0; i < t; i++)
    {
        int p = x[i].first, pe = x[i].second;
        fac[0] = 1;//fac[i]:1~i之間不被p整除%p^e的所有數的乘積
        for(int j = 1; j <= pe; j++)
        {
            if(j % p == 0)
                fac[j] = fac[j - 1];
            else
                fac[j] = fac[j - 1] * j % pe;//注意是%pe不是p
        }
        //phi(x) = x*∏(i=1~n)(1-1/pi)
        //phipe = pe*(1-1/p);
        phipe = (pe / p) * (p - 1);
        for(int j = 0;j<T;j++)
        {
                                // b
                                //Ca mod (p^pe)
            //cout<<binom(a[j],b[j],p,pe)<<endl;
            ans[j] = (ans[j] + binom(a[j], b[j], p, pe) * pr[i]) % M;
        }
    }
    for(int j = 0; j < T; j++)
        cout<<ans[j]<<'\n';
    return 0;
}

莫比烏斯反演

積性函數

void prime(int n)
{
	p[1] = 1;
    for (int i = 2; i <= n; i++)
    {
        if (!p[i])    p[i] = i, pr[++tot] = i, pe[i] = i; 
        for (int j = 1; j <= tot && pr[j] * i <= n; j++) {
            p[pr[j] * i] = pr[j];
            if(p[i] == pr[j])   
            {
            	pe[i * pr[j]] = pe[i] * pr[j];
            	break;
            }
	        else
	        	pe[i * pr[j]] = pr[j];
        }
    }
}
void solve()
{       
	cin>>n;
	prime(n);
	f[1] = 1;	// !可能要改
	for(int i = 2; i <= n; i++)
	{
		if(i == pe[i])	f[i] = ...
		else f[i] = f[pe[i]] * f[i / pe[i]];
	}
    return;
}

$\varphi$

int p[N], pr[N / 5], tot;
ll phi[N], sphi[N];
void prime(int n)
{
    phi[1] = 1; p[1] = 1;
    for (int i = 2; i <= n; i++)
    {
        if (!p[i])    p[i] = i, pr[++tot] = i, phi[i] = i - 1; 
        for (int j = 1; j <= tot && pr[j] * i <= n; j++) {
            p[pr[j] * i] = pr[j];
            if(p[i] == pr[j])   
            {
                phi[i * pr[j]] = phi[i] * pr[j];
                break;
            }
            else
                phi[i * pr[j]] = phi[i] * (pr[j] - 1);
        }
    }
    for(int i = 1; i <= N - 10; i++)
        sphi[i] = sphi[i - 1] + phi[i];
}

$\mu$

int p[N], pr[N / 5], tot, mu[N];
ll smu[N];
void prime(int n)
{
	mu[1] = 1;
    for (int i = 2; i <= n; i++)
    {
        if (!p[i])    p[i] = i, pr[++tot] = i, mu[i] = -1; 
        for (int j = 1; j <= tot && pr[j] * i <= n; j++) {
            p[pr[j] * i] = pr[j];
            if(p[i] == pr[j])   
            { 
            	mu[i * pr[j]] = 0;
            	break;
            }
	        else
	        	mu[i * pr[j]] = -mu[i];
        }
    }
    for(int i = 1; i <= N - 10; i++)
		smu[i] = smu[i - 1] + mu[i];
}

杜教篩

杜教篩（Sum）模板

常用于求數論函數前綴和

杜教篩 = 整數分塊+迪利克雷卷積+線性篩

時間復雜度:$O(n^{\frac{2}{3}})$

公式：$g(1)S(n) = \sum_{i = 1}^{n}h(i)-\sum_{i = 2}^{n}g(i)S(\lfloor \dfrac{n}{i}\rfloor)$

對于函數$f$ 構造出來$h,g$滿足$f = h*g$

常見構造：

$e = \mu*1$，即$[n=1]=\sum_{d|n}\mu(d)$
$Id = \varphi*1$，即$n = \sum_{d|n}\varphi(d)$

const int  N  = 6e6 + 10;
typedef long long ll;

int cnt, pri[N], mu[N], sum1[N], phi[N];
bool vis[N];
ll sum2[N];
map<int, ll> sumphi;
map<int, int> summu;

void init(int n)
{
    phi[1] = mu[1] = 1;
    vis[0] = vis[1] = 1;
    for(int i = 2; i < n; i++)
    {
        if(!vis[i])
        {
            pri[++cnt] = i;
            mu[i] = -1; 
            phi[i] = i - 1;
        }
        for(int j = 1; j <= cnt && pri[j] * i <= n; j++)
        {
            vis[i * pri[j]] = 1;
            if(i % pri[j] == 0)
            {
            	mu[i * pri[j]] = 0;
                phi[i * pri[j]] = phi[i] * pri[j];
                break;
            }
            else 
            {
                mu[i * pri[j]] = -mu[i];
                phi[i * pri[j]] = phi[i] * (pri[j] - 1);
            }
        }
    }
    for(int i = 1; i < n; i++)
    {
    	sum1[i] = sum1[i - 1] + mu[i];
        sum2[i] = sum2[i - 1] + phi[i];
    }
}
ll dlsmu(int x)
{
    if(x < N)   return sum1[x];
    if(summu[x])    return summu[x];
    int &ans = summu[x];
    ans += 1;
    for(ll l = 2; l <= x; l++)
    {
        ll d = x / l;
        ll r = x / d;
        ans -= (r - l + 1) * dlsmu(d);
        l = r;
    }
    return ans;
}
ll dlsphi(int x)
{
    if(x < N)   return sum2[x];
    if(sumphi[x])   return sumphi[x];
    ll &ans = sumphi[x];
    ans += (1ll * x + 1) * x / 2;
    // x + 1 !!!在模板題中爆int
    for(ll l = 2; l <= x; l++)
    {
        ll d = x / l, r = x / d;
        ans -= (r - l + 1) * dlsphi(d);
        l = r;
    }
    return ans;
}
int main()
{
	ios::sync_with_stdio(false),cin.tie(0),cout.tie(0);
    int t, n;
    cin>>t;
    init(N - 10);
    while(t--)
    {
        cin>>n;
        cout<<dlsphi(n)<<" "<<dlsmu(n)<<"\n";
    }
    return 0;
}

大素數判斷 miller_rabin $O(s \log_2 ^ 3 n)$， $n \leq 2 ^ {54}$

typedef long long ll;
ll qmi(ll a, ll b, ll mod)
{
    ll ans = 1 % mod;
    while(b)
    {
        if(b & 1) ans = ans * a % mod;
        a = a * a % mod;
        b >>= 1;
    }
    return ans;
}

bool witness(ll a, ll n)
{
	ll u = n - 1;
	int t = 0;
	//2^t*u
	while((u & 1) == 0) u >>= 1, t++;
	ll x1, x2;
	x1 = qmi(a, u, n);	// a ^ u mod n
	for(int i = 1; i <= t; i++)
	{
		x2 = qmi(x1, 2, n);
		if(x2 == 1 && x1 != 1 && x1 != n - 1)	
			return true;
		x1 = x2;
	}
	if(x1 != 1)	return true;
	return false;
}

int miller_rabin(ll n, ll s)
{
	if(n < 2)	return 0;
	if(n == 2)	return 1;
	if(n % 2 == 0)	return 0;
	for(int i = 0; i < s && i < n; i++)
	{
		ll a = rand() % (n - 1) + 1;
		if(witness(a, n))	return 0;
	}
	return 1;
}


int main()
{
	ios::sync_with_stdio(false),cin.tie(0),cout.tie(0);
	int m;
	while(cin>>m)
	{
		int cnt = 0;
		for(int i = 1; i <= m; i++)
		{
			ll n;	cin>>n;
			int s = 50;
			cnt += miller_rabin(n, s);
		}
		cout<<cnt<<"\n";
	}
	return 0;
}

勾股數的神奇性質

性質1:較大的兩個數之和等于這組數最小值的平方。

勾股數只有兩種：$奇數^2+偶數^2 = 奇數^2,偶數^2+偶數^2 = 偶數^2$

性質2：一個正奇數(除了1)與兩個和等于此數的平方的連續正整數的一組勾股數。

比如一正奇數$n$，那么以$n$為最小值的一組勾股數：$n,(n^2-1)/2,(n^2+1)/2$

性質3：一個正偶數(除了0，2，4)，后兩數之和的二倍等于最小數平方。

比如一正偶數$n$，那么以$n$為最小值的一組勾股數：$n,(n^2/4)-1,(n^2/4)+1$

性質4：一組勾股數的倍數還是勾股數。

以上性質用于尋找勾股數，而其逆命題判斷勾股數時可能會有覆蓋不完全現象(如20，21，29)。

DP

01背包

    for(int i = 1; i <= n; i++)
        for(int j = m; j >= v[i]; j--)
            f[j] = max(f[j], f[j - v[i]] + w[i]);

完全背包

    for(int i = 1; i <= n; i++)
        for(int j = v[i]; j <= m; j++)
            f[j] = max(f[j], f[j - v[i]] + w[i]);

多重背包

$O(NMS)$

    for(int i = 1; i <= n; i++)
        for(int j = 0; j <= m; j++)
            for(int k = 0; k <= s[i] && k * v[i] <= j; k++)
                f[i][j] = max(f[i][j], f[i - 1][j - k * v[i]] + k * w[i]);

$O(NM\log S)$

    for(int i = 1; i <= n; i++){
        int v, w, l;
        cin >> v >> w >> l;
        for(int k = 1; l ; k *= 2){
            if(l >= k) l -= k;
            else k = l, l = 0;
            for(int j = m; j >= v * k; j--)
                f[j] = max(f[j], f[j - v * k] + w * k);
        }
    }

$O(NM)$ ver 1

int n, m, f[N];
int q[N][3];
void solve()
{
    cin>>n>>m;
    for(int i = 1; i <= n; i++)
    {
        int v, w, s;    cin>>v>>w>>s;
        for(int j = 0; j < v; j++)
        {
            int h = 1, t = 0;
            for(int p = j, x = 1; p <= m; p += v, x++)
            {
                int val = f[p] - x * w, cnt = x + s;
                for(; h <= t && q[t][1] <= val; t--);
                q[++t][1] = val, q[t][2] = cnt;
                f[p] = q[h][1] + x * w;
                for(; h <= t && q[h][2] == x; h++);
            }
        }
    }
}

$O(NM)$ ver 2

void solve()
{
    cin>>n>>m;
    for(int i = 1; i <= n; i++)
    {
        int v, w, s;
        cin>>v>>w>>s;
        memcpy(g, dp, sizeof dp);
        for(int j = 0; j < v; j++)
        {
            int hh = 0, tt = -1;
            for(int k = j; k <= m; k += v)
            {
                if(hh <= tt && q[hh] < k - s * v) hh++;
                while(hh <= tt && g[q[tt]] - (q[tt] - j) / v * w <= g[k] - (k - j) / v * w)
                    tt--;
                q[++tt] = k;
                dp[k] = g[q[hh]] + (k - q[hh]) / v * w;
            }
        }
    }
    cout<<dp[m]<<endl;
    return;
}

分組背包

const int N = 110;
int n, m;
int f[N], w[N][N], v[N][N], s[N];

void solve()
{
    cin>>n>>m;
    for(int i = 1; i <= n; i++)   
    {
        cin>>s[i];
        for(int j = 0; j < s[i]; j++)
            cin>>v[i][j]>>w[i][j];
    }
    for(int i = 1; i <= n; i++)
        for(int j = m; j >= 0; j--)
            for(int k = 0; k < s[i]; k++)
                if(v[i][k] <= j)
                    f[j] = max(f[j], f[j - v[i][k]] + w[i][k]);
    cout<<f[m];
}

最長公共子序列

    for(int i = 1; i <= n; i++)
        for(int j = 1; j <= m; j++)
        {
            f[i][j] = max(f[i - 1][j], f[i][j - 1]);
            if(a[i] == b[j])
                f[i][j] = max(f[i][j], f[i - 1][j - 1] + 1);
        }

最長上升子序列

$O(N^2)$

    for(int i = 1; i <= n; i++) 
    {
        f[i] = 1;
        for(int j = 1; j < i; j++)
            if(a[i] > a[j])
                f[i] = max(f[i], f[j] + 1);
    }
    ans = f[1];
    for(int i = 1; i <= n;i++)
        ans = max(ans, f[i]);

$O(N\log N)$

    int len = 0;
    for (int i = 0; i < n; i++)
    {
        int l = 0, r = len;
        while (l < r)
        {
            int mid = (l + r + 1) >> 1;
            if (a[i] > q[mid])  l=mid;
            else r = mid-1;
        }
        len = max(len, r + 1);
        q[r + 1] = a[i];
    }
    cout << len << endl;

樹上背包

$O(N^2)$ 點數和背包大小都為$N$

const int N = 2e3 + 10;
const int inf = 1 << 29;
int n, q;
int sz[N], w[N], f[N][N], tmp[N];
vector<int> e[N];
void dfs(int u)
{
    sz[u] = 0;
    for(auto v : e[u])
    {
        dfs(v);
        for(int i = 0; i <= sz[u] + sz[v]; i++)
            tmp[i] = -inf;
        for(int i = 0; i <= sz[u]; i++)
            for(int j = 0; j <= sz[v]; j++)
                tmp[i + j] = max(tmp[i + j], f[u][i] + f[v][j]);
        sz[u] += sz[v]; 
        for(int i = sz[u]; i >= 0; i--)
            f[u][i] = tmp[i];   
    }
    sz[u]++;
    for(int i = sz[u]; i >= 1; i--)
        f[u][i] = f[u][i - 1] + w[u];
}

$O(NM^2)$

const int N = 5e4 + 10;
const int M = 100;
const int inf = 1 << 29;
int n, q;
int sz[N], w[N], f[N][M + 10], tmp[N];
vector<int> e[N];

void dfs(int u)
{
    sz[u] = 0;
    for(auto v : e[u])
    {
        dfs(v);
        for(int i = 0; i <= sz[u] + sz[v]; i++)
            tmp[i] = -inf;
        for(int i = 0; i <= sz[u] && i <= M; i++)
            for(int j = 0; j <= sz[v] && i + j <= M; j++)
                tmp[i + j] = max(tmp[i + j], f[u][i] + f[v][j]);
        sz[u] += sz[v]; 
        for(int i = min(M,sz[u]); i >= 0; i--)
            f[u][i] = tmp[i];   
    }
    sz[u]++;
    for(int i = min(M, sz[u]); i >= 1; i--)
        f[u][i] = f[u][i - 1] + w[u];
}

數位DP

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
int a[20];
ll dp[20][state];
ll dfs(int pos, /*state變量*/, bool lead/*前導零*/, bool limit/*數位上界變量*/)
{
    if(pos == -1) return 1;
    if(!limit && !lead && dp[pos][state] != -1) return dp[pos][state];
    int up = limit ? a[pos] : 9;
    ll ans = 0;
    for(int i = 0; i <= up; i++)
    {
        if() ...
        else if()...
        ans += dfs(pos - 1, /*狀態轉移*/, lead && i == 0, limit && i == a[pos]); 
    }  
    if(!limit && !lead) dp[pos][state] = ans;
    return ans;
}
ll solve(ll x)
{
    int pos = 0;
    while(x)
    {
        a[pos++] = x % 10;
        x /= 10;
    }
    return dfs(pos - 1/*從最高位開始枚舉*/, /*一系列狀態 */, true, true);
}
int main()
{
    ll le, ri;
    memset(dp, -1, sizeof(dp));
    while(~scanf("%lld%lld", &le, &ri))
    {
        printf("%lld\n", solve(ri) - solve(le - 1));
    }
}

區間DP

 for(int len=2;len<=n;len++)
{
    for(int i=1;i+len-1<=n;i++)
    {
        int l=i,r=i+len-1;
        f[l][r]=1e8;
        for(int k=l;k<r;k++)
        {
            f[l][r]=min(f[l][r],f[l][k]+f[k+1][r]+s[r]-s[l-1]);
        }
    }
}

Data structure

單調棧

給定一個長度為 $N$ 的整數數列，輸出每個數左邊第一個比它小的數，如果不存在則輸出 $?1$
。

int main()
{
    cin >> n;
    for(int i = 0; i < n; i++)
    {
        cin >> x;
        while(tt && a[tt] >= x)    tt--;   //tt有，棧頂大于待入棧元素
        if (tt == 0) cout << "-1 ";         //第一個數，棧無
        else cout << a[tt] << " ";          //輸出所求
        a[++tt] = x;                        //入棧，不符合的話下一次while就清理掉了
    }
    return 0;
}

單調隊列

滑動窗口

int main()
{
    cin >> n>>k;
    for(int i = 0; i < n; i++)
        cin >> a[i];
    int hh = 0, tt = -1;
    for(int i = 0; i < n; i++)
    {
        if(hh <= tt && i - k + 1 > qq[hh]) hh++;//處理框框隊尾,框框滿了處理隊尾
        while(hh <= tt && a[qq[tt]] >= a[i])   tt--;   //隊首
        qq[++tt] = i;                           //隊列加入新下標       
        if(i >= k - 1) cout << a[qq[hh]]<<" "; //框框滿了
    }
    cout << endl;
    hh = 0, tt = -1;
    for(int i = 0; i < n; i++)
    {
        if(hh <= tt && i - k + 1 > qq[hh]) hh++;
        while(hh <= tt && a[qq[tt]] <= a[i])   tt--;
        qq[++tt] = i;
        if(i >= k - 1) cout << a[qq[hh]]<<" ";
    }
    return 0;
}

并查集

普通板子

int find(int x)
{
    if(fa[x] != x)      fa[x] = find(fa[x]);
    return fa[x];
}

封裝DSU

struct DSU {
    std::vector<int> p, siz;
    DSU(int n) : p(n + 1), siz(n + 1, 1) { std::iota(p.begin(), p.end(), 0); }
    int find(int x) {
        return p[x] == x ? x : p[x] = find(p[x]);
    }
    bool same(int x, int y) { return find(x) == find(y); }
    bool merge(int x, int y) {
        x = find(x);
        y = find(y);
        if (x == y) return false;
        siz[x] += siz[y];
        p[y] = x;
        return true;
    }
    int size(int x) { return siz[find(x)]; }
};

帶權并查集

typedef long long ll;
const int N = 201000;
int n, q, fa[N];
ll w[N];

int find(int x)
{
	if(x == fa[x])	return x;
	int p = fa[x];
	find(p);//先把fa[x]做一個路徑壓縮，假設現在的fa[p] =  q
	/*
		w[x] = a[x]-a[p]
		fa[p] = q , w[p]:a[p]-a[q]
		fa[x] = p , w[x]:a[x]-a[p]
		a[x] - a[p] + a[p] - a[q] = a[x]+a[p]
	*/
	w[x] = w[x] + w[p];
	return fa[x] = fa[p];
}

int main()
{
	cin>>n>>q;
	//w[i] = a[i]-a[fa[i]];
	for(int i = 1; i <= n; i++)	
		fa[i] = i, w[i] = 0;
	ll t = 0;
	for(int i = 0; i < q; i++)
	{	
		int op, l, r;
		cin>>op>>l>>r;
		l = (l + t) % n + 1;
		r = (r + t) % n + 1;
		if(op == 2)
		{
			if(find(l) != find(r))	continue;
			else cout<<w[l] - w[r]<<endl;
			t = abs(w[l] - w[r]);
		}
		else if(op == 1)
		{
			int x;	cin>>x;
			if(find(l) == find(r))	continue;
			/*
				w[fa[l]] = a[fa[l]]-a[fa[r]];
				a[l]-a[r] = x;
				w[l] = a[l]-a[fa[l]],w[r] = a[r]-a[fa[r]];
				a[l] - w[l] = a[fa[l]],a[r]-w[r] = a[fa[r]]
				w[fa[l]] = a[l]-w[l]-(a[r]-w[r])
			*/
			//先改權值再變父親
			w[fa[l]] = x - w[l] + w[r];
			fa[fa[l]] = fa[r];
		}
	}
	return 0;
}

哈希

單哈希板子

const int p = 9999971, base = 101;
int n , m, c[N + 10], ha[N + 10], hb[N + 10];
char a[N + 10], b[N + 10];
int main()
{
    cin>>n>>m;
    cin>>a + 1>>b + 1;
    c[0] = 1;
    for(int i = 1; i <= 200000; i++)
       c[i] = (c[i - 1] * base) % p;
    for(int i = 1; i <= n; i++)
        ha[i] = (ha[i - 1] * base + a[i] - 'a') % p;
    for(int i = 1; i <= m; i++)
        hb[i] = (hb[i - 1] * base + b[i] - 'a') % p;
    int ans = 0;
    for(int i = 1; i + m - 1 <= n; i++)
        if((ha[i + m - 1] - 1ll * ha[i - 1] * c[m] % p + p) % p == hb[m])
            ++ans;
    cout<<ans<<endl;
    return 0;
}

雙哈希封裝

typedef pair<long long, long long> pll;
struct DoubleStringHash
{
    vector<long long> h1, h2, w1, w2;
    long long base1 = 131, base2 = 13331;
    long long p1 = 1e9 + 7, p2 = 1e9 + 9;
    void init(string s) {
        int len = s.size();
        s = " " + s;
        h1.resize(len + 1), w1.resize(len + 1);
        h2.resize(len + 1), w2.resize(len + 1);
        h1[0] = 0, w1[0] = 1;
        h2[0] = 0, w2[0] = 1;
        for(int i = 1; i <= len; i++) {
            h1[i] = (h1[i - 1] * base1 + s[i]) % p1, w1[i] = w1[i - 1] * base1 % p1;
            h2[i] = (h2[i - 1] * base2 + s[i]) % p2, w2[i] = w2[i - 1] * base2 % p2;
        }
    }
    pll get(int l, int r) {
        return {(h1[r] - h1[l - 1] * w1[r - l + 1] % p1 + p1) % p1, (h2[r] - h2[l - 1] * w2[r - l + 1] % p2 + p2) % p2};
    }
};

Trie tree

普通板子

bool isend[N];
int n, m, nxt[N][26], cnt;
void solve()
{
    cin>>n;
    for(int i = 1; i <= n; i++)
    {
        string s;   cin>>s;
        int len = s.size(), now = 0;
        for(int j = 0; j < len; j++)
        {
            int x = s[j] - 'a';
            if(!nxt[now][x])
                nxt[now][x] = ++cnt;
                now = nxt[now][x];
        }
        isend[now] = true;
    }
    cin>>m;
    for(int i = 1; i <= m; i++)
    {
        string s;   cin>>s;
        int len = s.size(), now = 0;
        bool ok = true;
        for(int j = 0; j < len; j++)
        {
            int x = s[j] - 'a';
            if(!nxt[now][x])
            {
                ok = false;   
                break;
            }
            now = nxt[now][x];
        }
        if(ok)
            if(!isend[now])
                ok = false;
        if(ok)
            cout<<1<<endl;
        else cout<<0<<endl;
    }
    return;
}

封裝

struct trie 
{
    int nxt[500010][26], cnt;
    bool isend[500010];  // 該結點結尾的字符串是否存在

    void insert(string s) {  // 插入字符串
    int now = 0;
    int l = s.size();
    for(int i = 0; i < l; i++) 
    {
        int x = s[i] - 'a';
        if(!nxt[now][x]) nxt[now][x] = ++cnt;  // 如果沒有，就添加結點
            now = nxt[now][x];
        }
        isend[now] = 1;
    }

    bool find(string s) {  
        int now = 0;
        int l = s.size();
        for(int i = 0; i < l; i++) {
            int x = s[i] - 'a';
            if(!nxt[now][x]) return 0;
            now = nxt[now][x];
        }
        return isend[now];// 查找字符串是否出現過
        //return true;//是某個單詞的前綴 
    }
}tr;

void solve()
{
    int n;  cin>>n;
    for(int i = 1; i <= n; i++)
    {
        string s;   cin>>s;
        tr.insert(s);
    }
    cin>>n;
    for(int i = 1; i <= n; i++)
    {
        string s;   cin>>s;
        if(tr.find(s))
            cout<<1<<endl;
        else cout<<0<<endl;
    }
    return;
}

01

const int N = 2e5 + 10;

struct 
{
    int sz, s[2];
}seg[N * 32];

int root, tot = 0, q;

void insert(int x)
{
    int p = root;
    for(int j = 29; j >= 0; j--)
    {
        int w = (x >> j) & 1;
        seg[p].sz++;
        if(seg[p].s[w] == 0)  seg[p].s[w] = ++tot;
        p = seg[p].s[w];
    }
    seg[p].sz++;
}
void del(int x)
{
    int p = root;
    for(int j = 29; j >= 0; j--)
    {
        int w = (x >> j) & 1;
        seg[p].sz--;
        // if(seg[p].s[w] == 0)  seg[p].s[w] = ++tot;
        p = seg[p].s[w];
    }
    seg[p].sz--;
}

int query(int x, int k) //詢問 S 中有多少個數按位異或上 x 的結果小于 k
{
    int p = root;
    int ans = 0;
    for(int j = 29; j >= 0; j--)
    {
        int w = (x >> j) & 1; 
        int t = (k >> j) & 1;           
        if(w == 0 && t == 0)
        {
            p = seg[p].s[0];
        }
        else if(w == 0 && t == 1)
        {
            ans += seg[seg[p].s[0]].sz;
            p = seg[p].s[1];
        }
        else if(w == 1 && t == 0)
        {
            // ans += seg[seg[p].s[1]].sz;
            p = seg[p].s[1];
        }
        else if(w == 1 && t == 1)
        {
            ans += seg[seg[p].s[1]].sz;
            p = seg[p].s[0];            
        }

        // 異或第 k 小
        // if(seg[seg[p].s[w]].sz >= k)
        //     p = seg[p].s[w];
        // else 
        // {
        //     k -= seg[seg[p].s[w]].sz;
        //     p = seg[p].s[1 ^ w];
        //     ans = ans ^ (1 << j);
        // }
    }
    return ans;
}

void solve()
{   
    cin>>q;
    root = ++tot;
    for(int i = 1; i <= q; i++)
    {
        int opt;    cin>>opt;
        if(opt == 1)
        {
            int x;  cin>>x;
            insert(x);
        }
        else if(opt == 2)
        {
            int x;  cin>>x;
            del(x);
        }
        else if(opt == 3)
        {
            int x, k;  cin>>x>>k;
            cout<<query(x, k)<<'\n';
        }
    }
    return;
}

KMP $O(m+n)$

初始化next數組

char a[1000010]; // 文本串
char b[1000010]; // 模板串(將被匹配的子串)
int kmp_next[1000010]; // next數組

void getNext(int m){
	int j = 0;
	// 初始化next[0]的值
	kmp_next[0] = 0;
	for(int i=1; i<m; ++i){
		// 當這一位不匹配時，將j指向此位之前最大公共前后綴的位置
		while(j>0 && b[i]!=b[j]) j=kmp_next[j-1];
		// 如果這一位匹配，那么將j+1，繼續判斷下一位
		if(b[i]==b[j]) ++j;
		// 更新next[i]的值
		kmp_next[i] = j;
	}
}

1.使用KMP尋找匹配位置

例：給定兩個字符串 a , b ，求 b 是否為 a 的子串，并輸出 b 在 a 中第一次出現的位置。

int kmp(int n,int m){
	int i, j = 0;
	// 初始化位置p = -1
	int p = -1;
	// 初始化next數組
	getNext(m);
	for(i=0; i<n; ++i){
		// 當這一位不匹配時，將j指向此位之前最大公共前后綴的位置
		while(j>0 && b[j]!=a[i]) j=kmp_next[j-1];
		// 如果這一位匹配，那么將j+1，繼續判斷下一位
		if(b[j]==a[i]) ++j;
		// 如果是子串(m位完全匹配)，則更新位置p的值，并中斷程序
		if(j==m){
			p = i - m + 1;
			break;
		}
	}
	// 返回位置p的值
	return p;
}

2.使用KMP計算匹配次數

例：給定兩個字符串 a , b ，并輸出 b 在 a 中出現的次數。

int kmp(int n,int m){
    // 初始化答案數量res = 0
	int i, j = 0, res = 0;
	// 初始化next數組
	getNext(m);
	for(i=0; i<n; ++i){
		// 當這一位不匹配時，將j指向此位之前最大公共前后綴的位置
		while(j>0 && b[j]!=a[i]) j=kmp_next[j-1];
		// 如果這一位匹配，那么將j+1，繼續判斷下一位
		if(b[j]==a[i]) ++j;
		// 如果是子串(m位完全匹配)，則答案數量增加，res = res + 1
		if(j==m) ++res;
	}
	// 返回答案數量
	return res;
}

DFS序

int l[N], r[N], tot, tid[N];
vector<int> e[N];
template<class T>
struct BIT {
    T c[N];
    int size;
    void resize(int s) { size = s;}
    T query(int x) { // 1 ... x
        assert(x <= size);
        T s = 0;
        for (; x; x -= x & (-x)) {
            s += c[x];
        }
        return s;
    }

    void modify(int x, T s) { // a[x] += s
        assert(x != 0);
        for (; x <= size; x += x & (-x)) {
            c[x] += s;
        }
    } 
};
void dfs(int u,int fa)
{
    l[u] = ++tot;
    tid[tot] = u;
    for(int v : e[u])
    {
        if(v == fa)   continue;
        dfs(v, u);
    }
    r[u] = tot;
}

樹狀數組

普通模板

int n,q;
ll tr[N], a[N];
void modify(int x, ll s)
{
    for(; x <= n; x += x & -x)
        tr[x] += s;
}

ll query(int x)
{
    ll res = 0;
    for(; x; x -= x & -x)
        res += c[x];
    return res;
}

dls封裝

template<class T>
struct BIT {
    T c[N];
    int size;
    void resize(int s) { size = s;}
    T query(int x) { // 1 ... x
        assert(x <= size);
        T s = 0;
        for (; x; x -= x & (-x)) {
        s += c[x];
    }
    return s;
    }

    void modify(int x, T s) { // a[x] += s
        assert(x != 0);
        for (; x <= size; x += x & (-x)) {
            c[x] += s;
        }
    } 
};

BIT<ll> c1, c2;

掃描線

二維數點

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N = 201000;
vector<int>vx;
vector<array<int,4>>event;;
int n,q,m;
int a[N],ans[N];
int c[N];
int query(int x){//1...x
	ll s = 0;
	for(;x;x-=x&(-x))
		s += c[x];
	return s;
}
void modify(int x,int s){//a[x]+=s
	for(;x<=m;x+=x&(-x))
		c[x]+=s;
}
int main()
{
	ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
	cin>>n>>q;
	for(int i = 1;i<=n;i++)
	{
		int x,y;
		cin>>x>>y;
		vx.push_back(x);
		//y坐標，類型,x坐標
		event.push_back({y,0,x});
	}
	for(int i = 1;i<=q;i++)
	{
		int x1,x2,y1,y2;
		cin>>x1>>x2>>y1>>y2;
		//用容斥
		//y坐標，類型1-,2+,x坐標,哪一個詢問
		//0,1,2的設置其實還包含了希望哪個事件先發生,坐標一樣的話,我們希望先插入再查詢
		event.push_back({y2,2,x2,i});
		event.push_back({y1-1,2,x1-1,i});
		event.push_back({y2,1,x1-1,i});
		event.push_back({y1-1,1,x2,i});
	}
	sort(event.begin(), event.end());
	sort(vx.begin(), vx.end());
	//去重
	vx.erase(unique(vx.begin(), vx.end()),vx.end());
	m = vx.size();
	for(auto evt:event)
	{
		if(evt[1]==0){//插入
			int y = lower_bound(vx.begin(), vx.end(),evt[2])-vx.begin()+1;//樹狀數組是1base的
			modify(y,1);
		}
		else{//查詢<=它的最后一個位置，即第一個>它的位置-1
			int y = upper_bound(vx.begin(), vx.end(),evt[2])-vx.begin()-1+1;//樹狀數組是1base的
			int tmp = query(y);
			if(evt[1]==1)ans[evt[3]] -= tmp;
			else ans[evt[3]] += tmp;
		}
	}
	for(int i = 1;i<=q;i++)
		cout<<ans[i]<<'\n';
	return 0;
}

矩形面積并

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N = 201000;
vector<int>vx;
vector<array<int,4>>event;;
int n,q,m;
 
struct info
{
	int minv,mincnt;
};
 
info operator+(const info &l,const info &r)
{
	info a;
	a.minv = min(l.minv,r.minv);
	if(l.minv==r.minv)a.mincnt = l.mincnt + r.mincnt;
	else if(l.minv<r.minv)a.mincnt = l.mincnt;
	else a.mincnt = r.mincnt;
	return a;
}
 
struct node{
	int t;
	info val;
}seg[N*8];
 
void update(int id)
{
	seg[id].val = seg[id*2].val+seg[id*2+1].val;
}
 
void settag(int id,int t)
{
	seg[id].val.minv = seg[id].val.minv+t;
	seg[id].t = seg[id].t + t;
}
 
void pushdown(int id)
{
	if(seg[id].t!=0)
	{
		settag(id*2,seg[id].t);
		settag(id*2+1,seg[id].t);
		seg[id].t = 0;
	}
}
 
void build(int id,int l,int r)
{
	if(l==r)
		seg[id].val = {0,vx[r]-vx[r-1]};
	else 
	{
		int mid = (l+r)>>1;
		build(id*2,l,mid);
		build(id*2+1,mid+1,r);
		update(id);
	}
}
 
 
void modify(int id,int l,int r,int x,int y,ll t){
	if(l==x&&r==y)
	{
		settag(id,t);
		return;
	}
	int mid = (l+r)/2;
	pushdown(id);
	if(y<=mid) modify(id*2,l,mid,x,y,t);
	else if(x>mid) modify(id*2+1,mid+1,r,x,y,t);
	else{
		modify(id*2,l,mid,x,mid,t),modify(id*2+1,mid+1,r,mid+1,y,t);
	}
	update(id);
}
 
 
int main()
{
	ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
	cin>>n;
	for(int i = 1;i<=n;i++)
	{
		int x1,x2,y1,y2;
		cin>>x1>>x2>>y1>>y2;
		vx.push_back(x1);
		vx.push_back(x2);
		//y坐標,類型0,x坐標
		event.push_back({y1,1,x1,x2});
		event.push_back({y2,-1,x1,x2});
	}
	sort(event.begin(), event.end());
	sort(vx.begin(), vx.end());
	//去重
	vx.erase(unique(vx.begin(), vx.end()),vx.end());
	m = vx.size()-1;//段數=點數-1
	build(1,1,m);
	int prey = 0;
	int totlen = seg[1].val.mincnt;
	ll ans = 0;
	for(auto evt:event)
	{
		int cov = totlen;
		if(seg[1].val.minv==0)
			cov = totlen - seg[1].val.mincnt;
		ans += (ll)(evt[0]-prey)*cov;
		prey = evt[0];
		int x1 = lower_bound(vx.begin(), vx.end(),evt[2])-vx.begin()+1;
		int x2 = lower_bound(vx.begin(), vx.end(),evt[3])-vx.begin();
		if(x1>x2)continue;
		modify(1,1,m,x1,x2,evt[1]);
	}
	cout<<ans<<endl;
	return 0;
}

區間不同數之和

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
#define int long long
const int N = 201000;
vector<int>vx;
vector<array<int,4>>event;;
int n,q,m;
int a[N],ans[N];
int c[N];
int pre[N],last[N];
int query(int x){//1...x
	ll s = 0;
	for(;x;x-=x&(-x))
		s += c[x];
	return s;
}
void modify(int x,int s){//a[x]+=s
	for(;x<=m;x+=x&(-x))
		c[x]+=s;
}
signed main()
{
	ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
	cin>>n>>q;
	m = n;
	for(int i = 1;i<=n;i++)
	{
		cin>>a[i];
		pre[i] = last[a[i]];//記錄一個數上一次出現的位置
		last[a[i]] = i;
		event.push_back({pre[i],0,i,a[i]});
	}
	/*
		對于一個區間[l,r]
		滿足條件的點需要：
		l<=i<=r
		pre[i]<l
	*/
	for(int i = 1;i<=q;i++)
	{
		int l,r;
		cin>>l>>r;
		event.push_back({l-1,1,l,i});
		event.push_back({l-1,2,r,i});
	}
	sort(event.begin(), event.end());//按pre[i]排序
	for(auto evt:event)
	{
		if(evt[1]==0){//插入
			modify(evt[2],evt[3]);
		}
		else{
			if(evt[1]==1)
				ans[evt[3]] -= query(evt[2]-1);
			else
				ans[evt[3]] += query(evt[2]);
		}
	}
	for(int i = 1;i<=q;i++)
		cout<<ans[i]<<'\n';
	 0;
}

線段樹

線段樹框架

struct info
{
    // ...

};

struct tag
{
    // ...

};

info operator + (const info &a, const info &b)
{
    info t;
    // ...

    return t;
}

struct segtree
{
    struct info val;
    struct tag tag;
    int sz;     // ...

}seg[N * 4];


void update(int id)
{
    // ...

}

void settag(int id, struct tag tag)
{
    // ...
    // 更新val，tag
    
}
void pushdown(int id)
{
    if(seg[id].tag != null)
    {
        settag(id * 2, seg[id].tag);
        settag(id * 2 + 1, seg[id].tag);
        seg[id].tag = null;
    }
}

void build(int id, int l, int r)
{
    seg[id].sz = r - l + 1;
    if(l == r)
    {

        return;
    }
    int mid = (l + r) >> 1;
    build(id * 2, l, mid);
    build(id * 2 + 1, mid + 1, r);
    update(id);
}
void change(int id, int l, int r, int pos, struct tag tag)
{
    if(l == r)
    {
        settag(id, tag);
        return;
    }
    pushdown(id);
    int mid = (l + r) >> 1;
    if(pos <= mid)
        change(id * 2, l, mid, pos, tag);
    else 
        change(id * 2 + 1, mid + 1, r, pos ,tag);
    update(id);
}

void modify(int id, int l, int r, int ql, int qr, struct tag tag)
{
    if(l == ql && r == qr)
    {
        settag(id, tag);
        return;
    }
    pushdown(id);
    int mid = (l + r) >> 1;
    if(qr <= mid)
        modify(id * 2, l, mid, ql, qr, tag);
    else if(ql > mid)
        modify(id * 2 + 1, mid + 1, r, ql, qr, tag);
    else 
        modify(id * 2, l, mid, ql, mid, tag),
        modify(id * 2 + 1, mid +1 , r, mid + 1, qr, tag);
    update(id);
}

info query(int id, int l, int r, int ql, int qr)
{
    if(ql == l && qr == r)
    {
        return seg[id].val;
    }
    pushdown(id);
    int mid = (l + r) >> 1;
    if(qr <= mid)
        return query(id * 2, l, mid, ql, qr);
    else if(ql > mid)
        return query(id * 2 + 1, mid + 1, r, ql, qr);
    else 
        return query(id * 2, l, mid, ql, mid) + query(id * 2 + 1, mid + 1, r, mid + 1, qr);
}

單點修改，區間查詢

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N = 201000;
int n,q;
int a[N];
 
struct node{
	int minv;
}seg[N*4];
 
void update(int id)
{
	seg[id].minv = min(seg[id*2].minv,seg[id*2+1].minv);
}
 
void build(int id,int l,int r)
{
	if(l==r)
		seg[id].minv = a[l];
	else 
	{
		int mid = (l+r)>>1;
		build(id*2,l,mid);
		build(id*2+1,mid+1,r);
		update(id);
	}
}
 
 
void change(int id,int l,int r,int pos,int val){
	if(l==r)
	{
		seg[id].minv = val;
		//a[pos] = val;已經把a數組記到線段樹里面了，之后不用了，可以不寫
	}
	else
	{
		int mid = (l+r)/2;
		if(pos<=mid)change(id*2,l,mid,pos,val);
		else change(id*2+1,mid+1,r,pos,val);
		//重要！改完之后記得更新節點的信息
		update(id);
	}
}
 
//O(logn)
int query(int id,int l,int r,int x,int y)
{
	if(l==x&&r==y)return seg[id].minv; 
	int mid = (l+r)/2;
	if(y<=mid)return query(id*2,l,mid,x,y);
	else if(x>mid)return query(id*2+1,mid+1,r,x,y);
	else{
		return min(query(id*2,l,mid,x,mid),query(id*2+1,mid+1,r,mid+1,y));
	}
}
 
 
int main()
{
	cin>>n>>q;
	for(int i = 1;i<=n;i++)
		cin>>a[i];
	build(1,1,n);
	for(int i = 1;i<=q;i++)
	{
		int op;
		cin>>op;
		if(op==1)
		{
			int x,d;
			cin>>x>>d;
			change(1,1,n,x,d);
		}
		else
		{
			int l,r;
			cin>>l>>r;
			cout<<query(1,1,n,l,r)<<endl;
		}
	}
	return 0;
}

區間加和區間最大值

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N = 201000;
int n,q;
int a[N];
 
struct node{
	ll t,val;
}seg[N*4];
 
void update(int id)
{
	seg[id].val = max(seg[id*2].val,seg[id*2+1].val);
}
 
void settag(int id,ll t)
{
	seg[id].val = seg[id].val+t;
	seg[id].t = seg[id].t + t;
}
 
void pushdown(int id)
{
	if(seg[id].t!=0)
	{
		settag(id*2,seg[id].t);
		settag(id*2+1,seg[id].t);
		seg[id].t = 0;
	}
}
 
void build(int id,int l,int r)
{
	if(l==r)
		seg[id].val = {a[l]};
	else 
	{
		int mid = (l+r)>>1;
		build(id*2,l,mid);
		build(id*2+1,mid+1,r);
		update(id);
	}
}
 
 
void modify(int id,int l,int r,int x,int y,ll t){
	if(l==x&&r==y)
	{
		settag(id,t);
		return;
	}
	int mid = (l+r)/2;
	pushdown(id);
	if(y<=mid) modify(id*2,l,mid,x,y,t);
	else if(x>mid) modify(id*2+1,mid+1,r,x,y,t);
	else{
		modify(id*2,l,mid,x,mid,t),modify(id*2+1,mid+1,r,mid+1,y,t);
	}
	update(id);
}
 
//O(logn)
ll query(int id,int l,int r,int x,int y)
{
	if(l==x&&r==y)return seg[id].val; 
	int mid = (l+r)/2;
	pushdown(id);
	if(y<=mid)return query(id*2,l,mid,x,y);
	else if(x>mid)return query(id*2+1,mid+1,r,x,y);
	else{
		return max(query(id*2,l,mid,x,mid),query(id*2+1,mid+1,r,mid+1,y));
	}
 
}
 
 
int main()
{
	cin>>n>>q;
	for(int i = 1;i<=n;i++)
		cin>>a[i];
	build(1,1,n);
	for(int i = 1;i<=q;i++)
	{
		int op;
		cin>>op;
		if(op==1)
		{
			int l,r;
			ll d;
			cin>>l>>r>>d;
			modify(1,1,n,l,r,d);
		}
		else
		{
			int l,r;
			cin>>l>>r;
			auto ans = query(1,1,n,l,r);
			cout<<ans<<endl;
		}
	}
	return 0;
}

線段樹維護最大字段和

ll a[N];
struct info
{
    ll ms,pre,suf,s;
};
info operator + (const info &l, const info &r)
{
    info a;
    a.s = l.s + r.s;
    a.ms = max({l.ms, r.ms, l.suf + r.pre});
    a.pre = max(l.pre, l.s + r.pre);
    a.suf = max(r.suf, r.s + l.suf);
    return a;
}
struct node
{
    info val;
} seg[N * 4];;

void update(int id)
{
    seg[id].val = seg[id * 2].val + seg[id * 2 + 1].val;
}

void build(int id,int l,int r)
{
    if (l == r)
    {
        seg[id].val = {a[l], a[l], a[l], a[l]};
    }
    else 
    {
        int mid = (l + r) >> 1;
        build(id * 2, l, mid);
        build(id * 2 + 1, mid + 1, r);
        update(id);
    }
}
void change(int id, int l, int r, int pos, int val)
{
    if (l == r)
    {
        seg[id].val = {a[l], a[l], a[l], a[l]};
    }
    else
    {
        int mid = (l + r) >> 1;
        if(pos <= mid)
            change(id * 2, l, mid, pos, val);
        else
            change(id * 2 + 1, mid + 1, r, pos, val);
        update(id);
    }
}
info query(int id, int l, int r, int ql, int qr)
{
    if (l == ql && r == qr)
        return seg[id].val;
    int mid = (l + r) >> 1;
    if (qr <= mid)
        return query(id * 2, l, mid, ql, qr);
    else if (ql > mid)
        return query(id * 2 + 1, mid + 1, r, ql, qr);
    else 
        return query(id * 2, l, mid, ql, mid) + query(id * 2 + 1, mid+1, r, mid + 1, qr);
}
int main()
{
    std::ios::sync_with_stdio(false);   cin.tie(nullptr), cout.tie(nullptr);
    int n,q;    cin>>n>>q;
    for(int i = 1; i <= n; i++)
        cin>>a[i];
    build(1, 1, n);
    while(q--)
    {
        int op,l,r; cin >> op >> l >> r;
        if (op == 1)
            change(1, 1, n, l, r);
        else if (op == 2)
        {
            auto ans = query(1, 1, n, l, r);
            cout<<ans.ms<<endl;
        }
    
    }
    return 0;
}

線段樹上二分

給$n$個數$a_1,a_2,a_3,…,a_n$。

支持$q$個操作：

1 x d，修改$ax=d$。
2 l r d，查詢$a_l,a_{l+1},a_{l+2},…,a_r$中大于等于$d$的第一個數的下標，如果不存在，輸出$?1$。也就是說，求最小的$i(l≤i≤r)$滿足$a_i≥d$。

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N = 201000;
int n,q;
int a[N];
 
struct node{
	int val;
}seg[N*4];
 
void update(int id)
{
	seg[id].val = max(seg[id*2].val,seg[id*2+1].val);
}
 
void build(int id,int l,int r)
{
	if(l==r)
		seg[id].val = a[l];
	else 
	{
		int mid = (l+r)>>1;
		build(id*2,l,mid);
		build(id*2+1,mid+1,r);
		update(id);
	}
}
 
 
void change(int id,int l,int r,int pos,int val){
	if(l==r)
	{
		seg[id].val = val;
		return;
	}
	int mid = (l+r)/2;
	if(pos<=mid)change(id*2,l,mid,pos,val);
	else change(id*2+1,mid+1,r,pos,val);
	update(id);
}
 
int search(int id,int l,int r,int x,int y,int d)
{
	if(l==x&&r==y)
	{
		if(seg[id].val<d)return -1;
		else
		{
			if(l==r)return l;
			int mid = (l+r)>>1;
			if(seg[id*2].val>=d)return search(id*2,l,mid,x,mid,d);
			else return search(id*2+1,mid+1,r,mid+1,y,d);
		}
	}
	int mid = (l+r)/2;
	if(y<=mid)return search(id*2,l,mid,x,y,d);
	else if(x>mid)return search(id*2+1,mid+1,r,x,y,d);
	else{
		int pos = search(id*2,l,mid,x,mid,d);
		if(pos==-1)
			return search(id*2+1,mid+1,r,mid+1,y,d);
		else 
			return pos; 
	}
}
 
int main()
{
	cin>>n>>q;
	for(int i = 1;i<=n;i++)
		cin>>a[i];
	build(1,1,n);
	for(int i = 1;i<=q;i++)
	{
		int op;
		cin>>op;
		if(op==1)
		{
			int x,d;
			cin>>x>>d;
			change(1,1,n,x,d);
		}
		else
		{
			int l,r,d;
			cin>>l>>r>>d;
			auto ans = search(1,1,n,l,r,d);
			cout<<ans<<endl;
		}
	}
	return 0;
}

動態開點線段樹

struct segtree
{
    int l, r;
    ll s, tag;
}seg[N * 30];
int n, m, tot = 1;

void update(int &id)
{
    seg[id].s = seg[seg[id].l].s + seg[seg[id].r].s;
}

void settag(int id, int sz, ll t)
{
    seg[id].s += t * sz;
    seg[id].tag += t;
}

void pushdown(int id, int l, int r)
{
    if(seg[id].tag == 0) return;
    if(seg[id].l == 0)  seg[id].l = ++tot;
    if(seg[id].r == 0)  seg[id].r = ++tot;
    int mid = (l + r) >> 1;
    if(seg[id].l != 0)
        settag(seg[id].l, mid - l + 1, seg[id].tag);
    if(seg[id].r != 0)
        settag(seg[id].r, r - mid, seg[id].tag);
    seg[id].tag = 0;
}

void change(int &id, int l, int r, int pos, int val)
{
    if(!id) 
    id = ++tot;
    if(l == r)
    {
        seg[id].s = val;
        return;
    }
    int mid = (l + r) >> 1;
    if(pos <= mid)
        change(seg[id].l, l, mid, pos, val);
    else
        change(seg[id].r, mid + 1, r, pos, val);
    update(id);
}

void modify(int &id, int l, int r, int ql, int qr, ll t)
{
    if(!id)
    id = ++tot;
    if(l == ql && r == qr)
    {
        settag(id, r - l + 1, t);
        return;
    }
    pushdown(id, l ,r);
    int mid = (l + r) >> 1;
    if(qr <= mid)
        modify(seg[id].l, l, mid, ql, qr, t);
    else if(ql > mid)   
        modify(seg[id].r, mid + 1, r, ql, qr, t);
    else
    {
        modify(seg[id].l, l, mid, ql, mid, t);
        modify(seg[id].r, mid + 1, r, mid + 1, qr, t);
    }
    update(id);
}

ll query(int id, int l, int r, int ql, int qr)
{
    if(!id)
    id = ++tot;
    if(l == ql && r == qr)
    {
        return seg[id].s;
    }
    pushdown(id, l, r);
    int mid = (l + r) >> 1;
    if(qr <= mid)
        return query(seg[id].l, l, mid, ql, qr);
    else if(ql > mid)   
        return query(seg[id].r, mid + 1, r, ql, qr);
    else
        return query(seg[id].l, l, mid, ql, mid) +
                query(seg[id].r , mid + 1, r, mid + 1, qr);
    }

可持久化線段樹第$k$小數

struct segtree
{
    int l, r, s;
}seg[N * 30];

vector<int> vx; 
int n, q, a[N], rt[N], tot;

void update(int id)
{
    seg[id].s = seg[seg[id].l].s + seg[seg[id].r].s;
}

int build(int l, int r)
{
    int id = ++tot;
    if(l == r)
        return id;
    int mid = (l + r) >> 1;
    seg[id].l = build(l, mid);
    seg[id].r = build(mid + 1, r);
    return id; 
}

int change(int u, int l, int r, int pos)
{
    int id = ++tot;
    seg[id] = seg[u];
    if(l == r)  
    {
        seg[id].s++;
        return id;
    }
    int mid = (l + r) >> 1;
    if(pos <= mid)
        seg[id].l = change(seg[u].l, l, mid, pos);
    else
        seg[id].r = change(seg[u].r, mid + 1, r, pos);
    update(id);
    return id;
}

int query(int v, int u, int l, int r, int x)
{
    if(l == r)
        return l;
    int cnt = seg[seg[u].l].s - seg[seg[v].l].s;
    int mid = (l + r) >> 1;
    if(x <= cnt)
        return query(seg[v].l, seg[u].l, l, mid, x);
    else
        return query(seg[v].r, seg[u].r, mid + 1, r, x - cnt);      
}

void solve()
{   
    cin>>n>>q;
    for(int i = 1; i <= n; i++)
    {
        cin>>a[i];
        vx.push_back(a[i]);
    }
    sort(vx.begin(), vx.end());
    vx.erase(unique(vx.begin(), vx.end()), vx.end());
    rt[0] = build(1, vx.size());
    for(int i = 1; i <= n; i++)
        rt[i] = change(rt[i - 1], 1, vx.size(), lower_bound(vx.begin(), vx.end(), a[i]) - vx.begin() + 1);

    while(q--)
    {
        int l, r, k;    cin>>l>>r>>k;
        cout<<vx[query(rt[l - 1], rt[r], 1, vx.size(), k) - 1]<<endl;
    }       
}

可持久化線段樹前$k$大之和

int n, m, q, tot, rt[N], a[N];
ll mi[N];
vector<int> vx;

struct segtree
{
    int l, r, cnt;
    ll s, val; 
}seg[N * 30];

void update(int id)
{
    seg[id].s = seg[seg[id].l].s + seg[seg[id].r].s;
    seg[id].cnt = seg[seg[id].l].cnt + seg[seg[id].r].cnt;
}

int build(int l, int r)
{
    int id = ++tot;
    if(l == r)
    {
    return id;
    }
    int mid = (l + r) >> 1;
    seg[id].l = build(l, mid);
    seg[id].r = build(mid + 1, r);
    update(id);
    return id;
}

int change(int u, int l, int r, int pos)
{
    int id = ++tot;
    seg[id] = seg[u];
    if(l == r)
    {
        seg[id].s = seg[id].s + vx[pos - 1];
        seg[id].cnt = seg[id].cnt + 1;
        seg[id].val = vx[l - 1];
        return id;
    }
    int mid = (l + r) >> 1;
    if(pos <= mid)
        seg[id].l = change(seg[id].l, l, mid, pos);
    else
        seg[id].r = change(seg[id].r, mid + 1, r, pos);
    update(id);
    return id;
}

ll query(int v, int u, int l, int r, int k)
{
    if(l == r)
    {
        return seg[u].val * k;
    }
    int mid = (l + r) >> 1;
    int tmp = seg[seg[u].r].cnt - seg[seg[v].r].cnt;
    if(tmp >= k)
        return query(seg[v].r, seg[u].r, mid + 1, r, k);
    else
        return seg[seg[u].r].s - seg[seg[v].r].s 
                + query(seg[v].l, seg[u].l, l, mid, k - tmp);
}

void  solve()
{
    tot = 0;
    vx.clear();

    cin>>n;
    for(int i = 1; i <= n; i++)
    {
        cin>>a[i];
        vx.push_back(a[i]);
    }

    sort(vx.begin(), vx.end());
    vx.erase(unique(vx.begin(), vx.end()), vx.end());   
    int m = vx.size();

    rt[0] = build(1, m);
    for(int i = 1; i <= n; i++)
    {
        int pos = lower_bound(vx.begin(), vx.end(), a[i]) - vx.begin() + 1;
        rt[i] = change(rt[i - 1], 1, m, pos);
    }

    cin>>q;
    for(int i = 1; i <= q; i++)
    {
        int l, r, k;    cin>>l>>r>>k;
        int mm = r - l + 1;
        int x = l, y = l + mm - 1;
        ll ans = query(rt[x - 1], rt[y], 1, m, k) + mi[mm];
        cout<<ans<<'\n'; 
    }
}

樹鏈剖分點權

//  AC one more times
ll mod, w[N];
vector<int> e[N];
int n, root, q;
int tot, l[N], r[N], tid[N], top[N], bs[N], sz[N], f[N], dep[N];

void dfs1(int u, int fa)
{
    sz[u] = 1;
    dep[u] = dep[fa] + 1;
    bs[u] = -1;
    f[u] = fa;
    for(auto v : e[u])
    {
        if(v == fa) continue;
        dfs1(v, u);
        sz[u] +=sz[v];
        if(bs[u] == -1 || sz[v] > sz[bs[u]])
            bs[u] = v;
    }
}

void dfs2(int u,int t)
{
    l[u] = ++tot;
    tid[tot] = u;
    top[u] = t;
    if(bs[u] != -1)
        dfs2(bs[u], t);
    for(auto v : e[u])
    {
        if(v == bs[u] || v == f[u])    continue;
        dfs2(v, v);
    }
    r[u] = tot;
}

struct segtree
{
    ll s, tag, sz;
}seg[N * 4];

void update(int id)
{
    seg[id].s = seg[id * 2].s + seg[id * 2 + 1].s;
    seg[id].s %= mod;
    seg[id].sz = seg[id * 2].sz + seg[id * 2 + 1].sz;
}

void settag(int id, ll tag)
{
    seg[id].tag += tag;
    seg[id].s += tag * seg[id].sz, seg[id].s %= mod;
}

void pushdown(int id)
{
    if(seg[id].tag != 0)
    {
    	settag(id * 2, seg[id].tag);
    	settag(id * 2 + 1, seg[id].tag);
    	seg[id].tag = 0;
    }
}

void build(int id, int l, int r)
{
    seg[id].sz = r - l + 1;
    if(l == r)
    {
        seg[id].s = w[tid[l]];
    	return;
    }
    int mid = (l + r) >> 1;
    build(id * 2, l, mid);
    build(id * 2 + 1, mid + 1, r);
    update(id);
}

void modify(int id, int l, int r, int ql, int qr, ll tag)
{
    if(l > r)   return;
    if(l == ql && r == qr)
    {
        settag(id, tag);
        return;
    }
    pushdown(id);
    int mid = (l + r) >> 1;
    if(qr <= mid)
        modify(id * 2, l, mid, ql, qr, tag);
    else if(ql > mid)
        modify(id * 2 + 1, mid + 1, r, ql, qr, tag);
    else 
    {
        modify(id * 2, l, mid, ql, mid, tag);
        modify(id * 2 + 1, mid + 1, r, mid + 1, qr, tag);
    }
    update(id);
}

ll query(int id, int l, int r, int ql, int qr)
{
    if(l > r)   return 0;
    if(l == ql && r == qr)
    {
        return seg[id].s;
    }
    pushdown(id);
    int mid = (l + r) >> 1;
    if(qr <= mid)
        return query(id * 2, l, mid, ql, qr);
    else if(ql > mid)
        return query(id * 2 + 1, mid + 1, r, ql, qr);
    else 
    {
        return (query(id * 2, l, mid, ql, mid) + 
                query(id * 2 + 1, mid + 1, r, mid + 1, qr)) % mod;
    }
    update(id);
}

void modify(int u, int v, ll k)
{
    while(top[u] != top[v])
    {
        if(dep[top[u]] > dep[top[v]])
        {
            modify(1, 1, n, l[top[u]], l[u], k);
            u = f[top[u]];
        }
        else
        {
            modify(1, 1, n, l[top[v]], l[v], k);
            v = f[top[v]];
        }
    }
    if(dep[u] < dep[v]) swap(u, v);
    modify(1, 1, n, l[v], l[u], k);
}

ll query(int u, int v)
{
    ll ans = 0;
    while(top[u] != top[v])
    {
        if(dep[top[u]] > dep[top[v]])
        {
            ans += query(1, 1, n, l[top[u]], l[u]);
            ans %= mod;
            u = f[top[u]];
        }
        else
        {
            ans += query(1, 1, n, l[top[v]], l[v]);
            ans %= mod;
            v = f[top[v]];
        }
    }
    if(dep[u] < dep[v]) swap(u, v);
    ans += query(1, 1, n, l[v], l[u]);
    ans %= mod;
    return ans;
}

void solve()
{
    cin>>n>>q>>root>>mod;
    for(int i = 1; i <= n; i++)
        cin>>w[i];
    for(int i = 1; i < n; i++)
    {
        int u, v;    cin>>u>>v;
        e[u].push_back(v);
        e[v].push_back(u);
    }
    dfs1(root, 0);
    dfs2(root, root);
    build(1, 1, n);
    while(q--)
    {
        int op; cin>>op;
        if(op == 1)
        {
            int u, v;   cin>>u>>v;
            ll k;   cin>>k;
            modify(u, v, k);
        }
        else if(op == 2)
        {
            int u, v;   cin>>u>>v;
            cout<<query(u, v) % mod<<endl;
        }
        else if(op == 3)
        {
            int u;  ll k;
            cin>>u>>k;
            modify(1, 1, n, l[u], r[u], k);
        }
        else if(op == 4)
        {
            int u;  cin>>u;
            cout<<query(1, 1, n, l[u], r[u]) % mod<<endl;
        }
    }
    return;
}

樹鏈剖分邊權下放

vector<PII> e[N];
int n, q;
int tot, l[N], r[N], tid[N], top[N], bs[N], sz[N], f[N], dep[N];
ll w[N];
pii edge[N];
void dfs1(int u, int fa)
{
    sz[u] = 1;
    dep[u] = dep[fa] + 1;
    bs[u] = -1;
    f[u] = fa;
    for(auto v : e[u])
    {
        if(v.fi == fa) continue;
        dfs1(v.fi, u);
        w[v.fi] = v.se;
        sz[u] +=sz[v.fi];
        if(bs[u] == -1 || sz[v.fi] > sz[bs[u]])
            bs[u] = v.fi;
    }
}

void dfs2(int u,int t)
{
    l[u] = ++tot;
    tid[tot] = u;
    top[u] = t;
    if(bs[u] != -1)
        dfs2(bs[u], t);
    for(auto v : e[u])
    {
        if(v.fi == bs[u] || v.fi == f[u])    continue;
            dfs2(v.fi, v.fi);
    }
    r[u] = tot;
}

struct info 
{
    ll s, mav, miv;
};

info operator + (const info &a, const info &b)
{
    return (info){a.s + b.s, max(a.mav, b.mav), min(a.miv, b.miv)};
}

struct segtree
{
    struct info val;
    int tag;
}seg[N * 4];

void update(int id)
{
    seg[id].val = seg[id * 2].val + seg[id * 2 + 1].val;
}

void settag(int id, int tag)
{
    seg[id].tag += tag, seg[id].tag %= 2;
    swap(seg[id].val.mav, seg[id].val.miv);
    seg[id].val = {-seg[id].val.s, -seg[id].val.mav, -seg[id].val.miv};
}

void pushdown(int id)
{
    if(seg[id].tag == 1)
    {
        settag(id * 2, seg[id].tag);
        settag(id * 2 + 1, seg[id].tag);
        seg[id].tag = 0;
    }
}

void build(int id, int l, int r)
{
    seg[id].tag = 0;
    if(l == r)
    {
        seg[id].val = {w[tid[l]], w[tid[l]], w[tid[l]]};
        return;
    }
    int mid = (l + r) >> 1;
    build(id * 2, l, mid);
    build(id * 2 + 1, mid + 1, r);
    update(id);
}

void change(int id, int l, int r, int pos, ll ew)
{
    if(l == r)
    {
        seg[id].val = {ew, ew, ew};
        return;
    }
    pushdown(id);
    int mid = (l + r) >> 1;
    if(pos <= mid)
        change(id * 2, l, mid, pos, ew);
    else if(pos > mid)
        change(id * 2 + 1, mid + 1, r, pos, ew);
    update(id);
}
void modify(int id, int l, int r, int ql, int qr, ll tag)
{
    if(ql > qr || l > r) return;
    if(l == ql && r == qr)
    {
        settag(id, tag);
        return;
    }
    pushdown(id);
    int mid = (l + r) >> 1;
    if(qr <= mid)
        modify(id * 2, l, mid, ql, qr, tag);
    else if(ql > mid)
        modify(id * 2 + 1, mid + 1, r, ql, qr, tag);
    else 
    {
        modify(id * 2, l, mid, ql, mid, tag);
        modify(id * 2 + 1, mid + 1, r, mid + 1, qr, tag);
    }
    update(id);
}

info query(int id, int l, int r, int ql, int qr)
{
    if((ql > qr || l > r)) return (info){0, -100000000, 100000000};
    if(l == ql && r == qr)
    {
        return seg[id].val;
    }
    pushdown(id);
    int mid = (l + r) >> 1;
    if(qr <= mid)
        return query(id * 2, l, mid, ql, qr);
    else if(ql > mid)
        return query(id * 2 + 1, mid + 1, r, ql, qr);
    else 
    {
        return query(id * 2, l, mid, ql, mid) +
            query(id * 2 + 1, mid + 1, r, mid + 1, qr);
    }
    update(id);
}

void modify(int u, int v, ll k)
{
    while(top[u] != top[v])
    {
        if(dep[top[u]] > dep[top[v]])
        {
            modify(1, 1, n, l[top[u]], l[u], k);
            u = f[top[u]];
        }
        else
        {
            modify(1, 1, n, l[top[v]], l[v], k);
            v = f[top[v]];
        }
    }
    if(dep[u] < dep[v]) swap(u, v);
    modify(1, 1, n, l[v] + 1, l[u], k);
}

info query(int u, int v)
{
    info ans = (info){0, -100000000, 100000000};
    while(top[u] != top[v])
    {
        if(dep[top[u]] > dep[top[v]])
        {
            ans = ans + query(1, 1, n, l[top[u]], l[u]);
            u = f[top[u]];
            }
        else
        {
            ans = ans + query(1, 1, n, l[top[v]], l[v]);
            v = f[top[v]];
        }
    }
    if(dep[u] < dep[v]) swap(u, v);
    ans = ans + query(1, 1, n, l[v] + 1, l[u]);
    return ans;
}

void solve()
{
    cin>>n;
    for(int i = 1; i < n; i++)
    {
        int u, v, w;    cin>>u>>v>>w;
        u++, v++;
        edge[i] = {u, v};
        e[u].push_bcak({v, w});
        e[v].push_bcak({u, w});
    }
    dfs1(1, 0);
    dfs2(1, 1);
    build(1, 1, n);
    cin>>q;
    while(q--)
    {
        string op;  cin>>op;
        if(op == "C")
        {
            int i, w;   cin>>i>>w;
            int u = edge[i].fi, v = edge[i].se;
            if(dep[u] < dep[v]) swap(u, v);
            change(1, 1, n, l[u], w);
        }
        else if(op == "N")
        {
            int u, v;
            cin>>u>>v;
            u++, v++;
            modify(u, v, 1);
        }
        else if(op == "SUM")
        {
            int u, v;
            cin>>u>>v;
            u++, v++;
            cout<<query(u, v).s<<endl;            
        }
        else if(op == "MAX")
        {
            int u, v;
            cin>>u>>v;
            u++, v++;
            cout<<query(u, v).mav<<endl;            
        }
        else if(op == "MIN")
        {
            int u, v;
            cin>>u>>v;
            u++, v++;
            cout<<query(u, v).miv<<endl;            
        }
    }
}

ST表

const int LOGN = 21;
const int N = 100000;
int f[N + 10][LOGN + 2], LOG[N + 10];
int n, m;
void pre()
{
    LOG[1] = 0, LOG[2] = 1;
    for(int i = 3; i <= N; i++)
        LOG[i] = LOG[i / 2] + 1;
    for(int j = 1; j <= LOGN; j++)
    {
        for(int i = 1; i + (1 << j) - 1 <= n; i++)
            f[i][j] = max(f[i][j - 1], f[i + (1 << (j - 1))][j - 1]);
    }
}
void solve()
{
    cin>>n>>m;
    for(int i = 1; i <= n; i++)   cin>>f[i][0];
    
    pre();

    for(int i = 1; i <= m; i++)
    {
        cin>>x>>y;
        int s = LOG[y - x + 1];
        cout<<max(f[x][s], f[y - (1 << s) + 1][s])<<endl;
    }
    return;
}

ST 表封裝

const int N = 2e5 + 10;
const int LOGN = 20;    // 2 ^ 20 = 1048576
struct S_T {
    // op 函數需要支持兩個可以重疊的區間進行合并
    // 例如 min、 max、 gcd、 lcm 等
    int f[LOGN + 2][N], lg[N];
    void build(int n) {
        lg[0] = -1;
        for(int i = 1; i <= n; ++i) {
            f[0][i] = ; // 手動指定原始數組
            lg[i] = lg[i / 2] + 1;
        }

        for(int i = 1; i <= LOGN; ++i)
            for(int j = 1; j + (1 << i) - 1 <= n; ++j)
                f[i][j] = max(f[i - 1][j], f[i - 1][j + (1 << (i - 1))]);
    }
    int query(int l, int r) {
        int len = lg[r - l + 1];
        return max(f[len][l], f[len][r - (1 << len) + 1]);
    }
}ST;

DSU ON TREE

const int N = 1e5 + 10;
int n, k;
vector<int> e[N];
int l[N], r[N], id[N], sz[N], hs[N], tot;

inline void add(int u)
{
    
}

inline void del(int u)
{
    
}

inline void query(int u)
{
    
}

void dfs_init(int u,int f) {
    l[u] = ++tot;
    id[tot] = u;
    sz[u] = 1;
    hs[u] = -1;
    for (auto v : e[u]) {
        if (v == f) continue;
        dfs_init(v, u);
        sz[u] += sz[v];
        if (hs[u] == -1 || sz[v] > sz[hs[u]])
            hs[u] = v;
    }
    r[u] = tot;
}

void dfs_solve(int u, int f, bool keep) {
    for (auto v : e[u]) {
        if (v != f && v != hs[u]) {
            dfs_solve(v, u, false);
        }
    }
    if (hs[u] != -1) {
        dfs_solve(hs[u], u, true);
    }

    for (auto v : e[u]) {
        if (v != f && v != hs[u]) {
            for (int x = l[v]; x <= r[v]; x++)
                query(id[x]);
            for (int x = l[v]; x <= r[v]; x++)
                add(id[x]);
        }
    }
    //query(u); 
    add(u);
    if (!keep) {
        for(int x = l[u]; x <= r[u]; x++)
            del(id[x]);
    }
}

void solve()
{       
    cin>>n;
    for (int i = 1; i < n; i++) {
        int u, v;   cin>>u>>v;
        e[u].push_back(v);
        e[v].push_back(u);
    }
    dfs_init(1, 0);
    dfs_solve(1, 0, false);
}

LCA

#include<bits/stdc++.h>
using namespace std;
const int N = 5000010;
const int LOGN = 20;
int n, m, root, dep[N], fa[N][LOGN + 2];
vector<int> e[N];
void dfs(int u, int from)
{
    dep[u] += dep[from] + 1;
    for(auto v : e[u]) {
        if(v == from) continue;
        fa[v][0] = u;
        dfs(v, u);
    }
}
void lca_init()
{
    for(int j = 1; j <= LOGN; j++)
        for(int i = 1; i <= n; i++)
            fa[i][j] = fa[fa[i][j - 1]][j - 1];
}
int lca_query(int u, int v)
{
    if(dep[u] < dep[v])   swap(u, v);
    int d = dep[u] - dep[v];
    for(int j = LOGN; j >= 0; j--)
        if(d & (1 << j))
            u = fa[u][j];
    if(u == v)    return u;
    for(int j = LOGN; j >= 0; j--)
        if(fa[u][j] != fa[v][j])
            u = fa[u][j], v = fa[v][j];
    return fa[u][0];
}
int main()
{
    cin>>n>>m>>root;
    for(int i = 2; i <= n; i++)
    {
        int u, v;    cin>>u>>v;
        e[u].push_back(v);
        e[v].push_back(u);
    }
    dfs(root, 0);
    lca_init();
    while(m--)
    {
        int u, v;    cin>>u>>v;
        cout<<lca_query(u, v)<<endl;
    }
return 0;
}

笛卡爾樹

int n, a[N], l[N], r[N], ans[N], cnt;
void dfs(int x)
{
	ans[x] = ++cnt;
	if(l[x])	dfs(l[x]);
	if(r[x])	dfs(r[x]);
}

void build()
{
	stack<int> st;
	int root = 0;
	for(int i = 1; i <= n; i++)
	{
		int last = 0;//在棧里面最后一個被彈出的節點
		while(!st.empty() && a[st.top()] > a[i])
		{
			last = st.top();
			st.pop();
		}
		if(!st.empty())	r[st.top()] = i;//新加入的元素設為棧頂元素的右兒子
		else root = i;
		l[i] = last;
		st.push(i);
	}
	// for(int i = 1;i<=n;i++)
	// 	cout<<"i = "<<i<<" a[i] = "<<a[i]<<" l[i] = "<<l[i]<<" r[i] = "<<r[i]<<endl;
	dfs(root);
}

int main()
{
	cin>>n;
	for(int i = 1;i<=n;i++)
		cin>>a[i];
	build();
	for(int i = 1;i<=n;i++)
		cout<<ans[i]<<" ";
	cout<<'\n';
}

Graph

樹的重心

int idx, ans = N, n;
vector<int> e[N];

int dfs(int u, int fa)
{
    int size = 0, sum = 0;
    for(auto v : e[u])
    {
        if(v == fa)  continue;
        int s = dfs(v, u);
        size = max(size, s);
        sum += s;
    }
    size = max(size, n - sum - 1);
    ans = min(ans, size);
    return sum + 1;
}
//    cout<<ans<<endl;

拓撲排序

int n, m, d[N], l[N], tot;
vector<int> e[N];
void toposort()
{
    queue<int> q;
    for(int i = 1; i <= n; i++)
        if(d[i] == 0)
            q.push(i);
    while(!q.empty())
    {
        int u = q.front();  q.pop();
        l[++tot] = u;
        for(auto v : e[u])
            if(--d[v] == 0)
                q.push(v);
    }   
    for(int i = 1; i <= n; i++)
        if(d[i] > 0)
        {
            cout<<"Yes"<<endl;   // 是否存在toposort
            //輸出toposort序列
            for(int j = 1; j <= tot; j++)
                cout<<l[j]<<" ";
            cout<<endl;
            return;
        }
    cout<<"No"<<endl;
}

void solve()
{   
    cin>>n>>m;
    for(int i = 1; i <= m; i++)
    {
        int x , y;    cin>>x>>y;
        son[x].push_back(y);
        d[y]++;
    }
    toposort();
    return;
}

歐拉圖

定義
通過圖中所有邊恰好一次的通路稱為歐拉通路。
通過圖中所有邊恰好一次的回路稱為歐拉回路。
具有歐拉回路的無向圖或有向圖稱為歐拉圖。
具有歐拉通路但不具有歐拉回路的無向圖或有向圖稱為半歐拉圖。
有向圖也可以有類似的定義。
非形式化地講，歐拉圖就是從任意一個點開始都可以一筆畫完整個圖，半歐拉圖必須從某個點開始才能一筆畫完整個圖。
性質
歐拉圖中所有頂點的度數都是偶數。
若 G是歐拉圖，則它為若干個環的并，且每條邊被包含在奇數個環內。
判別法
對于無向圖G ， G是歐拉圖當且僅當 G 是連通的且沒有奇度頂點。
對于無向圖G ， G是半歐拉圖當且僅當 G是連通的且 G 中恰有 0個或 2個奇度頂點。
對于有向圖 G， G是歐拉圖當且僅當 G的所有頂點屬于同一個強連通分量且每個頂點的入度和出度相同。
對于有向圖 G， G是半歐拉圖當且僅當
? 如果將G 中的所有有向邊退化為無向邊時，那么 G的所有頂點屬于同一個連通分量。
? 最多只有一個頂點的出度與入度差為 1。
? 最多只有一個頂點的入度與出度差為 1。
? 所有其他頂點的入度和出度相同。

有向圖

//單詞接龍
#include<bits/stdc++.h>
using namespace std;

std::vector<int> e[27];
int n, m, l, f[27], c[100002], ind[27], outd[27], v[27];

inline void dfs(int x)
{
    while(f[x] < v[x])  //當前邊存在
    {
        int y = e[x][f[x]];
        ++f[x];
        dfs(y);
        c[++l] = y;
    }
}

inline void Euler()
{
    int x = 0, y = 0, z = 0;//x起點，y有多少入度比出度大一，有多少入度不等于出度
    for(int i = 1;i <= n; i++)
    {
        if(outd[i] == ind[i] + 1)   y++, x = i;
        if(outd[i] != ind[i])   z++;
    }
    if(!(!z || (y == 1 && z == 2)))
    {
        cout<<"No\n";
        return;
    }
    if(!x)
        for(int i = 1 ; i <= n; i++)
            if(ind[i])
                x = i;
    memset(f, 0, sizeof(f));//表示看到當前的第幾條邊
    l = 0;
    dfs(x);
    c[++l] = x;
    if(l == m + 1)  cout<<"Yes\n";//注意要判斷連通性,l==m+1,說明經過了m條邊
    else cout<<"No\n";
}

int main()
{
    n = 26;
    cin>>m;
    for(int i = 1; i <= m; i++)
    {
        char str[101];
        cin>>str;
        int x = str[0]- 'a' + 1, y = str[strlen(str) - 1] - 'a' + 1;
        e[x].push_back(y);
        ++v[x]; //這里的v[x]其實就是outd,記錄每個點連出去多少條邊
        ++outd[x];
        ++ind[y];
    }
    Euler();
    return 0;
}

無向圖是否存在歐拉路

#include<bits/stdc++.h>
using namespace std;
const int N = 1e5 + 10;
struct Node{
    int y, idx;
    Node(int _y, int _idx){ y = _y, idx = _idx; };
};
vector<Node> e[N];
int n, m, cnt = 1, l, f[N], c[N], d[N], v[N];
bool b[N * 2];

inline void dfs(int x)
{
    while(f[x] < v[x])//當前邊存在
    {
        int y = e[x][f[x]].y, idx = e[x][f[x]].idx;
        if(!b[idx])
        {
            ++f[x];
            b[idx] = b[idx ^ 1]=true;
            dfs(y);
            c[++l] = y;
        }
        else ++f[x];
    }
}

inline void Euler()
{
    int x = 0, y = 0;//記錄度數為奇數點的個數
    for(int i = 1 ; i <= n; i++)
       if(d[i] & 1)
            ++y, x = i;

    if(y && y != 2)
    {
        cout<<"No\n";
        return;
    }
    if(!x)
        for(int i = 1;i <= n; i++)
            if(d[i])
                x = i;
    memset(b, false, sizeof(b));
    memset(f, 0, sizeof(f));
    l = 0;
    dfs(x);
    c[++l] = x;
    if(l != m + 1)
        cout<<"No\n";
    else
        cout<<"Yes\n";
}

int main()
{
    ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
    cin>>n>>m;
    for(int i = 1; i <= m; i++)
    {
        int u, v;
        cin>>u>>v;
        e[u].push_back({v, ++cnt});
        e[v].push_back({u, ++cnt});
        ++d[u];
        ++d[v];
    }
    for(int i = 1; i <= n; i++)
        v[i] = e[i].size();
    Euler();
    return 0;
}

分層圖

分層圖
一般模型：
在圖上,有 $k$ 次機會可以直接通過一條邊,問起點與終點之間的最短路徑.
時間復雜度 $O(Mk \log (Nk))$
注意：
1.編號為 $i$ 的點在第 $j$ 層的編號是 $i+j*n$
2.記得數組開4倍
3.層與層之間單向邊

const int N = 1e5 + 10;
int n, m, k, s, t;
vector<pair<int, int>> e[N * 4];
int dist[N * 4];
bool vis[N * 4];
void dijkstra(int s)
{
    priority_queue<pair<int,int>, vector<pair<int, int>>, greater<pair<int, int>>> q;
    memset(dist, 0x3f, sizeof dist);
    dist[s] = 0;
    q.push({0, s});
    while(q.size() >= 1)
    {
        auto t = q.top();       q.pop();
        int u = t.second;
        if(vis[u])   continue;
        vis[u] = true;
        for(auto v : e[u])
        {
            if(dist[v.first] > dist[u] + v.second)
            {
                dist[v.first] = dist[u] + v.second;
                q.push({dist[v.first], v.first});
            }
        }
    }
}

int main()
{
	ios::sync_with_stdio(false),cin.tie(0),cout.tie(0);
	cin>>n>>m>>k;
	cin>>s>>t;
	s++, t++;
	for(int i = 1; i <= m; i++)
	{
		int u, v, w;	cin>>u>>v>>w;
		u++, v++;
		e[u].push_back({v, w});
		e[v].push_back({u, w});
		for(int j = 1;j<=k;j++)
		{
			//層于層之間對應點連一條權值為0的邊
			e[(u + (j - 1) * n)].push_back({v + (j * n), 0});
			e[(v + (j - 1) * n)].push_back({u + (j * n), 0});
			//每一層對應點的連邊
			e[(u + j * n)].push_back({v + j * n, w});
			e[(v + j * n)].push_back({u + j * n, w});
		}
	}
	dijkstra(s);
	int ans = 1e9;
	for(int i = 0; i <= k; i++)
		ans = min(ans, dist[t + i * n]);
	cout<<ans<<'\n';
	return 0;
}

Dijkstra

$O(N\log M)$

int n, m, dist[N];
vector<pair<int, int>> e[N];
bool vis[N];
void dijkstra(int start)
{
    priority_queue<pair<int,int>, vector<pair<int, int>>, greater<pair<int, int>>> q;
    memset(dist, 0x3f, sizeof dist);
    memset(vis, false, sizeof vis);
    dist[start] = 0;
    q.push({0, start});
    while(q.size() >= 1)
    {
        auto t = q.top();       q.pop();
        int u = t.second;
        if(vis[u])   continue;
        vis[u] = true;
        for(auto [v, w] : e[u])
        {
            if(dist[v] > dist[u] + w)
            {
                dist[v] = dist[u] + w;
                q.push({dist[v], v});
            }
        }
    }
}

$O(N^2)$

int n, m, g[N][N], dist[N];
bool st[N];
int dijkstra()
{
    memset(dist, 0x3f, sizeof dist);
    dist[1] = 0;
    for(int i = 0; i < n - 1; i++)
    {
        int t = -1;
        for(int j = 1; j <= n; j++)
        {
            if (!st[j] && (t == -1 || dist[t] > dist[j]))
                t = j;
        }
        for(int j = 1; j <= n; j++)
            dist[j] = min(dist[j], dist[t] + g[t][j]);
        st[t] = true;
    }
    if(dist[n] == 0x3f3f3f3f) return -1;
    return dist[n];
}

void solve()
{   
    cin >> n >> m;
    memset(g, 0x3f, sizeof g);
    while(m--)
    {
        int a, b, c;
        cin >> a >> b >> c;
        g[a][b] = min(g[a][b], c);
    }
    cout << dijkstra();
    return;
}

Bellman-ford $O(NM)$

int n, m, k, dist[N], backup[N];
struct EDGES
{
    int a, b, w;
}edge[M];
int bellman_ford()
{
    memset(dist, 0x3f, sizeof dist);
    dist[1] = 0;
    for(int i = 0; i < k; i++)
    {
        memcpy(backup, dist, sizeof dist);
        for(int j = 0; j < m; j++)
        {
            auto e = edge[j];
            dist[e.b] = min(dist[e.b], backup[e.a] + e.w);
        }
    }
}

void solve()
{   
    cin >> n >> m >> k;
    for(int i = 0; i < m; i++)
    {
        int u, v, w;
        cin >> u >> v >> w;
        edge[i] = {u, v, w};
    }
    bellman_ford();
    return;
}

SPFA 判負環

const int N = 4e5 + 10, M = 1e4 + 10;
bool in[N];
int cnt[N],h[N];
inline bool spfa_check_negative_ring(int s, int n)
{
	memset(h, 127, sizeof(h));
	memset(in, false, sizeof(in));
	memset(cnt, 0, sizeof(cnt));
	queue<int> q;
	h[s] = 0;
	in[s] = true;
	cnt[s] = 1;
	q.push(s);
	while(!q.empty())
	{
		int u = q.front();	q.pop();
		in[u] = false;
		for(auto [v, w]: edge[u])
		{	
			if(h[u] + w < h[v])
			{
				h[v] = h[u] + w;
				if(!in[v])
				{
					q.push(v);
					cnt[v]++;
					in[v]=true;
					if(cnt[v]>=n)
					{
						return true;
					}
				}
			}
		}
	}
	return false;
}

SPFA 次短路

更新了最小距離，要把上次的最小距離先拿給次小距離，因為上次的最小距離就是比當前距離大且最小的距離（即為次短距離）。
雖然可能當前路徑無法更新最小距離，但可能更新次小距離，要加判斷
從上一個點的次短距離更新這一個點的次短距離

const int N = 10100;
int n, m;
vector<pair<int, int>> edge[N];
int dist[2][N], c[N];
bool in[5001];//是否在隊列里面
int sum;
inline void spfa(int s)
{
    memset(dist, 127, sizeof(dist));
    memset(in, false, sizeof(in));
    queue<int> q;
    dist[0][s] = 0;
    dist[1][s] = 0;
    in[s] = true;
    q.push(s);
    while(!q.empty()){
        int u = q.front();
        q.pop();
        in[u] = false;
        for(auto [v, w] : edge[u])
        {
            if(dist[0][u] + w < dist[0][v])
            {   //更新最短路
                dist[1][v] = dist[0][v];
                dist[0][v] = dist[0][u] + w;
                if(!in[v]){
                    q.push(v);
                    in[v] = true;
                }
            }
            if((dist[1][v] > dist[0][u] + w && dist[0][u] + w > dist[0][v]) ||
                dist[1][v] == dist[0][v])
            {   //當前不能更新最短路，但可以更新次短路
                dist[1][v] = dist[0][u] + w;
                if(!in[v]){
                    q.push(v);
                    in[v] = true;
                }
            }
            if(dist[1][v] > dist[1][u] + w && dist[1][u] + w > dist[0][v])
            {   //用次短路更新次短路
                dist[1][v] = dist[1][u] + w;
                if(!in[v]){
                    q.push(v);
                    in[v] = true;
                }
            }
        }
    }
    return;
}
int main()
{
    ios::sync_with_stdio(false),cin.tie(0),cout.tie(0);
    cin>>n>>m;
    for(int i = 1; i <= m; i++)
    {
        int u, v, w;
        cin>>u>>v>>w;
        edge[u].push_back({v, w});
        edge[v].push_back({u, w});
    }
    spfa(1);
    cout<<dist[1][n]<<"\n";
    return 0;
}

Floyd $O(N^3)$

int d[N][N];
void floyd()
{
    for(int k = 1; k <= n; k++)
        for(int i = 1; i <= n; i++)
            for(int j = 1; j <= n; j++)
                d[i][j] = min(d[i][j], d[i][k] + d[k][j]);
}

void solve()
{   
    cin >> n >> m;
    for(int i = 1; i <= n; i++)
        for(int j = 1; j <= n; j++)
            if (i == j) d[i][j] = 0;
            else d[i][j] = INF;
    for(int i = 1; i <= m; i++)
    {
        int u, v, w;
        cin >> u >> v >> w;
        d[u][v] = min(d[u][v], w);
    }
    floyd();    
    return;
}

Johnson 全源最短路稀疏圖$O(NM \log M)$ 最慢 $O((N + M) \times (C + 1))$，$C$ 是環數

#include<bits/stdc++.h>
using namespace std;

typedef long long ll;
const int INF = 1e9;
const int N = 3e3 + 10, M = 1e4 + 10;
// 邊數是 N + M

vector<pair<int, int>>edge[M];
int n, m, k;
int c[N], cnt[N];
ll dist[N], h[N];
bool in[N];


bool vis[N];
void dijkstra(int start)
{
	priority_queue<pair<ll, int>, vector<pair<ll, int>>, greater<pair<ll, int>>> q;
	memset(vis, false, sizeof vis);
	for(int i = 0; i <= n; i++)
		dist[i] = INF;
	dist[start] = 0;
	q.push({0, start});
	while(q.size() >= 1)
	{
		auto t = q.top();       q.pop();
		int u = t.second;
		if(vis[u])   continue;
		vis[u] = true;
		for(auto [v, w] : edge[u])
		{
			if(dist[v] > dist[u] + w)
			{
				dist[v] = dist[u] + w;
				q.push({dist[v], v});
			}
		}
	}
}

inline bool spfa_check_negative_ring(int s, int n)
{
	memset(h, 127, sizeof(h));
	memset(in, false, sizeof(in));
	memset(cnt, 0, sizeof(cnt));
	queue<int> q;
	h[s] = 0;
	in[s] = true;
	cnt[s] = 1;
	q.push(s);
	while(!q.empty())
	{
		int u = q.front();	q.pop();
		in[u] = false;
		for(auto [v, w]: edge[u])
		{	//x連出去的邊i
			if(h[u] + w < h[v])
			{
				h[v] = h[u] + w;
				if(!in[v])
				{
					q.push(v);
					cnt[v]++;
					in[v]=true;
					if(cnt[v]>=n)
					{
						return true;
					}
				}
			}
		}
	}
	return false;
}
int main()
{
	ios::sync_with_stdio(false),cin.tie(0),cout.tie(0);
	cin>>n>>m;
	for(int i = 1; i <= m; i++)
	{
		int u, v, w;
		cin>>u>>v>>w;
		edge[u].push_back({v, w});
	}
	for(int i = 1; i <= n; i++)
		edge[0].push_back({i, 0});
	if(spfa_check_negative_ring(0, n))
	{
		cout<<-1<<"\n";
		return 0;
	}
	for(int u = 1; u <= n; u++)
	{
		for(auto &[v, w] : edge[u])
		{
			w += h[u] - h[v];
		}
	}
	for(int i = 1; i <= n; i++)
	{
		dijkstra(i);
		ll ans = 0;
		for(int j = 1; j <= n; j++)
		{
			if(dist[j] == INF)
				ans += 1ll * j * INF;
			else ans += 1ll * j * (dist[j] + h[j] - h[i]);
		}
		cout<<ans<<"\n";
	}
	return 0;
}

最小生成樹

Kruskal $O(M\log M)$

int n, m, p[N];
struct Edge
{
    int u, v;
    ll w;
    bool operator <(const Edge& W)const
    {
        return w < W.w;
    }
}edge[M];
int find(int x)
{
    if (p[x] != x)  p[x] = find(p[x]);
    return p[x];
}
void kruskal()
{
    sort(edge + 1, edge + 1 + m);
    ll res = 0;
    for (int i = 1; i <= n; i++)    p[i] = i;
        for (int i = 1; i <= m; i++)
        {
            int u = edge[i].u, v = edge[i].v;
            ll w = edge[i].w;
            int fu = find(u), fv = find(v);
            if (fu != fv)
            {
                p[fu] = fv;
                res += w;
            }
        }
}

int main()
{
    cin >> n >> m;
    for (int i = 1; i <= m; i++)
    {
        int u, v;	ll w;    cin >> u >> v >> w;
        edge[i] = {u, v, w};
    }
    kruskal();

}

Kruskal 重構樹

修改點邊數量點數開兩倍 kruskal_Rebuild_Tree 同樣

#include<bits/stdc++.h>
#define INF 0x3f3f3f3f
using namespace std;

typedef long long ll;
const int N = 4e5 + 10, M = 5e5+10;
const int LOGN = 20;
struct Node
{
	int u, v, w;
}a[M + 1];

int n, m, q, now, fa[N][LOGN + 2], dep[N];
vector<int> e[N];
int ls[N], rs[N], val[N];
ll sum[N], c[N], dp[N];

//////////////////////////////DSU//////////////////////////////////////
struct Dsu {
	int fa[N];
	void init(int x) {for(int i = 1; i <= x; i++) fa[i] = i;}
	int find(int x) {return x == fa[x] ? x : fa[x] = find(fa[x]);}
	void merge(int x, int y) {
		int u = find(x), v = find(y);
		if(u == v) return;
		fa[u] = v;
	}
}dsu;
/////////////////////////Kruskal////////////////////////////////////

void kruskalRebuildTree()
{
	dsu.init(2 * n - 1);
	sort(a + 1, a + 1 + m, [](const Node& x, const Node& y) {
		//return x.w > y.w;	//最大生成樹(最小值最大)
		return x.w < y.w; 	//最小生成樹(最大值最小)
	});
	now = n;
	for(int i = 1; i <= m; i++) {
		ll u = dsu.find(a[i].u), v = dsu.find(a[i].v), w = a[i].w;
		if(u != v) {
			val[++now] = w;
			dsu.merge(u, now);
			dsu.merge(v, now);
//			e[now].push_back(u);
//			e[now].push_back(v);
			ls[now] = u;
			rs[now] = v;
		}
	}
}


////////////////////////////DP///////////////////////////////////

void dfs(int u)
{
	if(!ls[u] || !rs[u])
	{
		sum[u] = c[u];
		dp[u] = INF;
		return;
	}
	dfs(ls[u]), dfs(rs[u]);
	sum[u] = sum[ls[u]] + sum[rs[u]];
	//               先吃ls[u]
	dp[u] = max(min(dp[rs[u]], val[u]) - sum[ls[u]], 
		min(dp[ls[u]], val[u]) - sum[rs[u]]);
}

////////////////////////////Main///////////////////////////////////
int main()
{
	cin>>n>>m;
	for(int i = 1; i <= n; i++)	cin>>c[i];
	for(int i = 1; i <= m; i++)
		cin>>a[i].u>>a[i].v>>a[i].w;
	kruskalRebuildTree();
	dfs(now);
	return 0;
}

Prim $O(N^2)$，稀疏圖可以用堆優化$O(M\log N)$，但Kruskal更好

int n, m, g[N][N], dist[N];
bool st[N];
int prim()
{
    memset(dist, 0x3f, sizeof dist);
    int res = 0;
    for(int i = 0; i < n; i++)
    {
        int t = -1;
        for (int j = 1; j <= n; j++)
        {
            if(st[j] == false && (t == -1 || dist[t] > dist[j]))  
                t = j;
        }
        if(i && dist[t] == INF) return INF;
        if(i)  res += dist[t];
        st[t] = true;
        for (int j = 1; j <= n; j++)    
            dist[j] = min(dist[j], g[t][j]);
    }
    return res;
}

int main()
{
    memset(g, 0x3f, sizeof g);
    cin >> n >> m;
    for(int i = 1; i <= m; i++)
    {
        int u, v, c;
        cin >> u >> v >> c;
        g[u][v] = g[v][u] = min(g[u][v], c);
    }
    int t = prim();
    if(t == INF)   cout << "impossible";
    else    cout << t;
    return 0;
}

二分圖

二分圖染色 $O(N + M)$

#include<iostream>
#include<cstring>
#include<vector>
using namespace std;

vector<int> e[1010];
int n, m, c[1010];
bool dfs(int u)
{
    for(auto v : e[u])
    {
        if(c[v] == 0)
        {
            c[v] = 3 - c[u];
            if(!dfs(v))
                return false;
        }
        else if(c[u] == c[v])
        return false;
    }
    return true;
}
bool check()
{
    memset(c, 0 ,sizeof c);
    for(int i = 1; i <= n; i++)
    {
        if(c[i] == 0)
        {
            c[i] = 1;
            if(!dfs(i))
                return false;
        }     
    }
    return true;
}
int main()
{
    cin>>n>>m;
    for(int i = 1; i <= m; i++)
    {
        int u, v;    cin>>u>>v;
        e[u].push_back(v);
        e[v].push_back(u);
    }
    if(check())
        cout<<"Yes"<<endl;
    else cout<<"No"<<endl;
    return 0;
}

二分圖最大匹配 $O(NM)$

const int N = 2e6 + 10;

int n, n1, n2, m, v[N];
int b[N];
vector<int> e[N];
int now;
bool find(int x)
{
    b[x] = now;
    for(auto &y : e[x])
    {
        if(!v[y] || (b[v[y]] != now && find(v[y])))
        {
        	b[v[y]] = now;
            v[y] = x;
            return true;
        }
    }
    return false;
}
int match()
{
    memset(v, 0, sizeof v);
    for(int i = 1; i < 1e6 + 10;i++)
    {
       // memset(b,0,sizeof(b));
    	++now;
        if(!find(i))
        {
        	return i-1;
        }
    }
    return 0;
}
int main()
{
	ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
    cin>>n;
    n1 = n;
    // 建圖 染色


    cout<<match()<<endl;
    return 0;
}

連通性相關

SCC Trajan

vector<int> edge[N];
int n, m;
int dfn[N], low[N], ins[N], idx, bel[N], cnt, sz[N];//sz：每個強連通分量的size
//          low記這個子樹里面能跳到的dfn最小的，且未被切掉的(ins[u] = true)
stack<int> stk;
void dfs(int u)
{
	dfn[u] = low[u] = ++idx;
	ins[u] = true;
	stk.push(u);//還沒被切掉的點
	for(auto v : edge[u])
	{
		if(!dfn[v])	dfs(v);
		if(ins[v])	low[u] = min(low[u], low[v]);
	}
	if(dfn[u] == low[u]){
		++cnt;
		while(1)
		{
			int v = stk.top();	stk.pop();
			ins[v] = false;
			bel[v] = cnt;
			sz[cnt]++;
			if(u == v)break;
		}
	}
}

void tarjan()
{
	for(int i = 1; i <= n; i++)
		if(!dfn[i])	dfs(i);
}

int main()
{
	ios::sync_with_stdio(false),cin.tie(0),cout.tie(0);

	cin>>n>>m;
	for(int i = 1; i <= m; i++)
	{
		int u, v;	cin>>u>>v;
		edge[u].push_back(v);
	}
	tarjan();
}

SCC Kosaraju

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;

const int N = 101000;

vector<int> edge[N], erev[N];
vector<int> out;
int n, m;
bool vis[N];

void dfs(int u)
{
	vis[u] = true;
	for(auto v : edge[u])
	{
		if(!vis[v])	dfs(v);
	}
	out.push_back(u);
}


void dfs2(int u)
{
	vis[u] = true;
	for(auto v : erev[u])
	{
		if(!vis[v])	dfs2(v);
	}
}


void kosaraju()
{
	for(int i = 1; i <= n; i++)
	{
		if(!vis[i])	dfs(i);
	}
	reverse(out.begin(), out.end());
	memset(vis, false, sizeof(vis));
	for(auto u : out)
	{
		if(!vis[u])
		{
			dfs2(u);
		}
	}
}

int main()
{
	ios::sync_with_stdio(false),cin.tie(0),cout.tie(0);
	cin>>n>>m;
	for(int i = 1; i <= m; i++)
	{
		int u, v;	cin>>u>>v;
		edge[u].push_back(v);
		erev[v].push_back(u);
	}
	kosaraju();
}

縮點

vector<int> edge[N];
int n, m;
int dfn[N], low[N], ins[N], idx, bel[N], cnt, sz[N];//sz：每個強連通分量的size
//         low記這個子樹里面能跳到的dfn最小的，且未被切掉的(ins[u] = true)
stack<int> stk;
ll ans, dp[N];
int a[N];
void dfs(int u)
{
	dfn[u] = low[u] = ++idx;
	ins[u] = true;
	stk.push(u);//還沒被切掉的點
	for(auto v : edge[u])
	{
		if(!dfn[v])	dfs(v);
		if(ins[v])	low[u] = min(low[u], low[v]);
	}
	if(dfn[u] == low[u]){
		++cnt;
		dp[cnt] = 0;
		ll sval = 0;
		while(1)
		{
			int x = stk.top();	stk.pop();
			ins[x] = false;
			bel[x] = cnt;
			sz[cnt]++;
			sval += a[x];
			for(auto y : edge[x])
			{
				if(bel[y] != cnt && bel[y] != 0)
					dp[cnt] = max(dp[cnt], dp[bel[y]]);
			}
			if(u == x)break;
		}
		dp[cnt] += sval;
		ans = max(ans, dp[cnt]);
	}
}

void tarjan()
{
	for(int i = 1; i <= n; i++)
	{
		if(!dfn[i])	dfs(i);
	}
}

int main()
{
	ios::sync_with_stdio(false),cin.tie(0),cout.tie(0);

	cin>>n>>m;
	for(int i = 1; i <= n; i++)
		cin>>a[i];
	for(int i = 1; i <= m; i++)
	{
		int u, v;	cin>>u>>v;
		edge[u].push_back(v);
	}
	tarjan();
	cout<<ans<<endl;
}

割點（圖不一定聯通）

const int N = 101000;
vector<int> edge[N];
int n, m, idx, sz;
int dfn[N], low[N], cut[N];
//         low記這個子樹里面能跳到的dfn最小的，且未被切掉的(ins[u] = true)


void dfs(int u, int fa, int root)
{
	dfn[u] = low[u] = ++idx;
	int ch = 0;//記兒子個數
	for(auto v : edge[u])
	{
		if(!dfn[v]) {
			dfs(v, u, root);
			ch++;
			low[u] = min(low[u], low[v]);
			if(low[v] >= dfn[u]) cut[u] = 1;//跳不出去了，說明它是割點
		}
		else if(v != fa)//返祖邊
		{
			low[u] = min(low[u], dfn[v]);//?這里就不能亂寫了，要對dfn取min了
		}
		
	}
	if(u == root && ch <= 1)	cut[u] = 0;
	sz += cut[u];
}

int main()
{
	cin>>n>>m;
	for(int i = 1; i <= m; i++)
	{
		int u, v;	cin>>u>>v;
		edge[u].push_back(v);
		edge[v].push_back(u);
	}
	for(int i = 1; i <= n; i++)
		if(!dfn[i])
			dfs(i, 0, i);
	cout<<sz<<"\n";
	for(int i = 1; i <= n; i++)
		if(cut[i])
			cout<<i<<"\n";
}

割邊

const int N = 101000;

vector<pair<int, int>> edge[N];
int n, m, idx;
int dfn[N], low[N];
vector<int>bridge;

void dfs(int u,int id)
{
	dfn[u] = low[u] = ++idx;
	for(auto [v, id2] : edge[u])
	{
		if(!dfn[v]){
			dfs(v, id2);
			low[u] = min(low[u], low[v]);
			if(dfn[v] == low[v])	bridge.push_back(id2);
		}
		else if(id != id2){
		//要看它是父親直接下來的邊，還是返祖邊，因為不能把樹邊當成返祖邊去更新low,那么這里只要不是父親就可以了
		//?注意：這里要寫它邊的編號不是它父親連下來的邊的編號，而不是說判這個點是不是父親
		//因為單純判是不是父親的話，會把重邊判成同一條邊，這樣是錯的。因為重邊是會影響是不是割邊
			low[u] = min(low[u], dfn[v]);
		}
	}
}

int main()
{
	cin>>n>>m;
	for(int i = 1; i <= m; i++)
	{
		int u, v;	cin>>u>>v;
		edge[u].push_back({v, i});
		edge[v].push_back({u, i});
	}

	dfs(1, -1);
	sort(bridge.begin(), bridge.end());
	cout<<bridge.size()<<"\n";
	for(auto x : bridge)
		cout<<x<<" ";
	cout<<endl;
}

(e-DCC)Tarjan邊雙縮點

const int N = 101000;
vector<pair<int, int>>edge[N];
int n, m;
int dfn[N], low[N], idx, bel[N], cnt;
stack<int> stk;
vector<int> cc[N];

void dfs(int u, int id)
{
	dfn[u] = low[u] = ++idx;
	stk.push(u);
	for(auto [v, id2] : edge[u])
	{
		if(!dfn[v]){
			dfs(v, id2);
			low[u] = min(low[u], low[v]);
		}
		else if(id != id2){
			low[u] = min(low[u], dfn[v]);
		}
	}
	if(dfn[u] == low[u]){
		++cnt;
		while(1)
		{
			int v = stk.top();
			cc[cnt].push_back(v);
			bel[v] = cnt;
			stk.pop();
			if(u == v)	break;
		}
	}

}

int main()
{
	cin>>n>>m;
	for(int i = 1; i <= m; i++)
	{
		int u, v;	cin>>u>>v;
		edge[u].push_back({v, i});
		edge[v].push_back({u, i});
	}
	dfs(1, -1);
	int nleaf = 0;
	for(int i = 1; i<= cnt; i++)
	{
		int cnte = 0;
		for(auto u : cc[i])
			for(auto [v, id] : edge[u])
				cnte += (bel[u] != bel[v]);
		nleaf += (cnte == 1);//度為1的點是葉子
	}
	cout<<(nleaf + 1) / 2<<"\n";
}

2 - SAT $O(\texttt{最短路算法})$

typedef long long ll;
const int N = 101000;

vector<int> edge[N];
int n, m;
int dfn[N], low[N], ins[N], idx, bel[N], cnt;
stack<int> stk;

void dfs(int u)
{
	dfn[u] = low[u] = ++idx;
	ins[u] = true;
	stk.push(u);
	for(auto v : edge[u])
	{
		if(!dfn[v])	dfs(v);
		if(ins[v])	low[u] = min(low[u], low[v]);
	}
	if(dfn[u] == low[u]){
		++cnt;
		while(1)
		{
			int v = stk.top();
			ins[v] = false;
			bel[v] = cnt;
			stk.pop();
			if(u == v)	break;
		}
	}

}
int main()
{
	ios::sync_with_stdio(false),cin.tie(0),cout.tie(0);
	int t;	cin>>t;
	while(t--)
	{
		cin>>n>>m;
		//標號0...2n-1 對于滿式mi:2*i,hi:2*i+1
		for(int i = 0; i < 2 * n; i++)
			dfn[i] = 0, edge[i].clear();
		idx = cnt = 0;
		for(int i = 1; i <= m; i++)
		{
			char ty1, ty2;
			int u, v;
			cin>>ty1>>u>>ty2>>v;
			--u, --v;
			u = u * 2 + (ty1 == 'h');
			v = v * 2 + (ty2 == 'h');
			edge[u ^ 1].push_back(v);
			edge[v ^ 1].push_back(u);
		}

		for(int i = 0; i < 2 * n; i++)
		{
			if(!dfn[i])
				dfs(i);
		}
		bool ok = true;
		for(int i = 0; i < n; i++)
		{
			// bel[2 * i] < bel[2 * i + 1]選 2 * i
			if(bel[2 * i]==bel[2 * i + 1])
			{
				ok = false;
				break;
			}
		}
		if(ok)
			cout<<"GOOD\n";
		else
			cout<<"BAD\n";
	}
}

計算幾何

模塊1

typedef double db;
const db EPS = 1e-9;
  
inline int sign(db a) { return a < -EPS ? -1 : a > EPS; }
  
inline int cmp(db a, db b){ return sign(a-b); }
  
struct P {
	db x, y;
	P() {}
	P(db _x, db _y) : x(_x), y(_y) {}
	P operator+(P p) { return {x + p.x, y + p.y}; }
	P operator-(P p) { return {x - p.x, y - p.y}; }
	P operator*(db d) { return {x * d, y * d}; }
	P operator/(db d) { return {x / d, y / d}; }

	bool operator<(P p) const { 
		int c = cmp(x, p.x);
		if (c) return c == -1;
		return cmp(y, p.y) == -1;
	}

	bool operator==(P o) const{
		return cmp(x,o.x) == 0 && cmp(y,o.y) == 0;
	}

	db dot(P p) { return x * p.x + y * p.y; }
	db det(P p) { return x * p.y - y * p.x; }
	 
	db distTo(P p) { return (*this-p).abs(); }
	db alpha() { return atan2(y, x); }
	void read() { cin>>x>>y; }
	void write() {cout<<"("<<x<<","<<y<<")"<<endl;}
	db abs() { return sqrt(abs2());}
	db abs2() { return x * x + y * y; }
	P rot90() { return P(-y,x);}
	P unit() { return *this/abs(); }
	int quad() const { return sign(y) == 1 || (sign(y) == 0 && sign(x) >= 0); }
	P rot(db an){ return {x*cos(an) - y*sin(an),x*sin(an) + y*cos(an)}; }
};

#define cross(p1,p2,p3) ((p2.x-p1.x)*(p3.y-p1.y)-(p3.x-p1.x)*(p2.y-p1.y))
#define crossOp(p1,p2,p3) sign(cross(p1,p2,p3))
 
// 直線 p1p2, q1q2 是否恰有一個交點
bool chkLL(P p1, P p2, P q1, P q2) {
	db a1 = cross(q1, q2, p1), a2 = -cross(q1, q2, p2);
	return sign(a1+a2) != 0;
}

// 求直線 p1p2, q1q2 的交點
P isLL(P p1, P p2, P q1, P q2) {
	db a1 = cross(q1, q2, p1), a2 = -cross(q1, q2, p2);
	return (p1 * a2 + p2 * a1) / (a1 + a2);
}

// 判斷區間 [l1, r1], [l2, r2] 是否相交
bool intersect(db l1,db r1,db l2,db r2){
	if (l1>r1) swap(l1,r1); if (l2>r2) swap(l2,r2); 
	return !( cmp(r1,l2) == -1 || cmp(r2,l1) == -1 );
}

// 線段 p1p2, q1q2 相交
bool isSS(P p1, P p2, P q1, P q2){
	return intersect(p1.x,p2.x,q1.x,q2.x) && intersect(p1.y,p2.y,q1.y,q2.y) && 
	crossOp(p1,p2,q1) * crossOp(p1,p2,q2) <= 0 && crossOp(q1,q2,p1)
			* crossOp(q1,q2,p2) <= 0;
}

// 線段 p1p2, q1q2 嚴格相交  
bool isSS_strict(P p1, P p2, P q1, P q2){
	return crossOp(p1,p2,q1) * crossOp(p1,p2,q2) < 0 && crossOp(q1,q2,p1)
			* crossOp(q1,q2,p2) < 0;
}

// m 在 a 和 b 之間
bool isMiddle(db a, db m, db b) {
	/*if (a > b) swap(a, b);
	return cmp(a, m) <= 0 && cmp(m, b) <= 0;*/
	return sign(a - m) == 0 || sign(b - m) == 0 || (a < m != b < m);
}

bool isMiddle(P a, P m, P b) {
	return isMiddle(a.x, m.x, b.x) && isMiddle(a.y, m.y, b.y);
}

// 點 p 在線段 p1p2 上
bool onSeg(P p1, P p2, P q){
	return crossOp(p1,p2,q) == 0 && isMiddle(p1, q, p2);
}
// q1q2 和 p1p2 的交點 在 p1p2 上？

// 點 p 嚴格在 p1p2 上
bool onSeg_strict(P p1, P p2, P q){
	return crossOp(p1,p2,q) == 0 && sign((q-p1).dot(p1-p2)) * sign((q-p2).dot(p1-p2)) < 0;
}

// 求 q 到 直線p1p2 的投影（垂足） ?? : p1 != p2
P proj(P p1, P p2, P q) {
	P dir = p2 - p1;
	return p1 + dir * (dir.dot(q - p1) / dir.abs2());
}

// 求 q 以 直線p1p2 為軸的反射
P reflect(P p1, P p2, P q){
	return proj(p1,p2,q) * 2 - q;
}

// 求 q 到 線段p1p2 的最小距離
db nearest(P p1, P p2, P q){
	if (p1 == p2) return p1.distTo(q);
	P h = proj(p1,p2,q);
	if(isMiddle(p1,h,p2))
		return q.distTo(h);
	return min(p1.distTo(q),p2.distTo(q));
}

// 求 線段p1p2 與 線段q1q2 的距離
db disSS(P p1, P p2, P q1, P q2){
	if(isSS(p1,p2,q1,q2)) return 0;
	return min(min(nearest(p1,p2,q1),nearest(p1,p2,q2)), min(nearest(q1,q2,p1),nearest(q1,q2,p2)));
}

// 極角排序

sort(p, p + n, [&](P a, P b) {
	int qa = a.quad(), qb = b.quad();
	if (qa != qb) return qa < qb;
	else return sign(a.det(b)) > 0;
});

模塊2

db area(vector<P> ps){
	db ret = 0; rep(i,0,ps.size()) ret += ps[i].det(ps[(i+1)%ps.size()]); 
	return ret/2;
}
  
int contain(vector<P> ps, P p){ //2:inside,1:on_seg,0:outside
	int n = ps.size(), ret = 0; 
	rep(i,0,n){
		P u=ps[i],v=ps[(i+1)%n];
		if(onSeg(u,v,p)) return 1;
		if(cmp(u.y,v.y)<=0) swap(u,v);
		if(cmp(p.y,u.y) > 0 || cmp(p.y,v.y) <= 0) continue;
		ret ^= crossOp(p,u,v) > 0;
	}
	return ret*2;
}
  
vector<P> convexHull(vector<P> ps) {
	int n = ps.size(); if(n <= 1) return ps;
	sort(ps.begin(), ps.end());
	vector<P> qs(n * 2); int k = 0;
	for (int i = 0; i < n; qs[k++] = ps[i++]) 
		while (k > 1 && crossOp(qs[k - 2], qs[k - 1], ps[i]) <= 0) --k;
	for (int i = n - 2, t = k; i >= 0; qs[k++] = ps[i--])
		while (k > t && crossOp(qs[k - 2], qs[k - 1], ps[i]) <= 0) --k;
	qs.resize(k - 1);
	return qs;
}
  
vector<P> convexHullNonStrict(vector<P> ps) {
	//caution: need to unique the Ps first
	int n = ps.size(); if(n <= 1) return ps;
	sort(ps.begin(), ps.end());
	vector<P> qs(n * 2); int k = 0;
	for (int i = 0; i < n; qs[k++] = ps[i++]) 
		while (k > 1 && crossOp(qs[k - 2], qs[k - 1], ps[i]) < 0) --k;
	for (int i = n - 2, t = k; i >= 0; qs[k++] = ps[i--])
		while (k > t && crossOp(qs[k - 2], qs[k - 1], ps[i]) < 0) --k;
	qs.resize(k - 1);
	return qs;
}
  
db convexDiameter(vector<P> ps){
	int n = ps.size(); if(n <= 1) return 0;
	int is = 0, js = 0; rep(k,1,n) is = ps[k]<ps[is]?k:is, js = ps[js] < ps[k]?k:js;
	int i = is, j = js;
	db ret = ps[i].distTo(ps[j]);
	do{
		if((ps[(i+1)%n]-ps[i]).det(ps[(j+1)%n]-ps[j]) >= 0)
			(++j)%=n;
		else
			(++i)%=n;
		ret = max(ret,ps[i].distTo(ps[j]));
	}while(i!=is || j!=js);
	return ret;
}
  
vector<P> convexCut(const vector<P>&ps, P q1, P q2) {
	vector<P> qs;
	int n = ps.size();
	rep(i,0,n){
		P p1 = ps[i], p2 = ps[(i+1)%n];
		int d1 = crossOp(q1,q2,p1), d2 = crossOp(q1,q2,p2);
		if(d1 >= 0) qs.push_back(p1);
		if(d1 * d2 < 0) qs.push_back(isLL(p1,p2,q1,q2));
	}
	return qs;
}

#include <bits/stdc++.h>
using namespace std;

typedef double db;
#define rep(i, a, n) for (int i = a; i < n; i++)
const db EPS = 1e-9;
  
inline int sign(db a) { return a < -EPS ? -1 : a > EPS; }

inline int cmp(db a, db b){ return sign(a-b); }
  
struct P {
	db x, y;
	P() {}
	P(db _x, db _y) : x(_x), y(_y) {}
	P operator+(P p) { return {x + p.x, y + p.y}; }
	P operator-(P p) { return {x - p.x, y - p.y}; }
	P operator*(db d) { return {x * d, y * d}; }
	P operator/(db d) { return {x / d, y / d}; }

	bool operator<(P p) const { 
		int c = cmp(y, p.y);
		if (c) return c == -1;
		return cmp(x, p.x) == -1;
	}

	bool operator==(P o) const{
		return cmp(x,o.x) == 0 && cmp(y,o.y) == 0;
	}

	db dot(P p) { return x * p.x + y * p.y; }
	db det(P p) { return x * p.y - y * p.x; }
	 
	db distTo(P p) { return (*this-p).abs(); }
	db alpha() { return atan2(y, x); }
	void read() {
		int x_, y_;
		scanf("%d%d", &x_, &y_);
		x = x_, y = y_;
	}

	void write() {cout<<"("<<x<<","<<y<<")"<<endl;}
	db abs() { return sqrt(abs2());}
	db abs2() { return x * x + y * y; }
	P rot90() { return P(-y,x);}
	P unit() { return *this/abs(); }
	int quad() const { return sign(y) == 1 || (sign(y) == 0 && sign(x) >= 0); }
	P rot(db an){ return {x*cos(an) - y*sin(an),x*sin(an) + y*cos(an)}; }
};

#define cross(p1,p2,p3) ((p2.x-p1.x)*(p3.y-p1.y)-(p3.x-p1.x)*(p2.y-p1.y))
#define crossOp(p1,p2,p3) sign(cross(p1,p2,p3))
 
// 直線 p1p2, q1q2 是否恰有一個交點
bool chkLL(P p1, P p2, P q1, P q2) {
	db a1 = cross(q1, q2, p1), a2 = -cross(q1, q2, p2);
	return sign(a1+a2) != 0;
}

// 求直線 p1p2, q1q2 的交點
P isLL(P p1, P p2, P q1, P q2) {
	db a1 = cross(q1, q2, p1), a2 = -cross(q1, q2, p2);
	return (p1 * a2 + p2 * a1) / (a1 + a2);
}

// 判斷區間 [l1, r1], [l2, r2] 是否相交
bool intersect(db l1,db r1,db l2,db r2){
	if (l1>r1) swap(l1,r1); if (l2>r2) swap(l2,r2); 
	return !( cmp(r1,l2) == -1 || cmp(r2,l1) == -1 );
}

// 線段 p1p2, q1q2 相交
bool isSS(P p1, P p2, P q1, P q2){
	return intersect(p1.x,p2.x,q1.x,q2.x) && intersect(p1.y,p2.y,q1.y,q2.y) && 
	crossOp(p1,p2,q1) * crossOp(p1,p2,q2) <= 0 && crossOp(q1,q2,p1)
			* crossOp(q1,q2,p2) <= 0;
}

// 線段 p1p2, q1q2 嚴格相交  
bool isSS_strict(P p1, P p2, P q1, P q2){
	return crossOp(p1,p2,q1) * crossOp(p1,p2,q2) < 0 && crossOp(q1,q2,p1)
			* crossOp(q1,q2,p2) < 0;
}

// m 在 a 和 b 之間
bool isMiddle(db a, db m, db b) {
	/*if (a > b) swap(a, b);
	return cmp(a, m) <= 0 && cmp(m, b) <= 0;*/
	return sign(a - m) == 0 || sign(b - m) == 0 || (a < m != b < m);
}

bool isMiddle(P a, P m, P b) {
	return isMiddle(a.x, m.x, b.x) && isMiddle(a.y, m.y, b.y);
}

// 點 p 在線段 p1p2 上
bool onSeg(P p1, P p2, P q){
	return crossOp(p1,p2,q) == 0 && isMiddle(p1, q, p2);
}
// q1q2 和 p1p2 的交點 在 p1p2 上？

// 點 p 嚴格在 p1p2 上
bool onSeg_strict(P p1, P p2, P q){
	return crossOp(p1,p2,q) == 0 && sign((q-p1).dot(p1-p2)) * sign((q-p2).dot(p1-p2)) < 0;
}

// 求 q 到 直線p1p2 的投影（垂足） ?? : p1 != p2
P proj(P p1, P p2, P q) {
	P dir = p2 - p1;
	return p1 + dir * (dir.dot(q - p1) / dir.abs2());
}

// 求 q 以 直線p1p2 為軸的反射
P reflect(P p1, P p2, P q){
	return proj(p1,p2,q) * 2 - q;
}

// 求 q 到 線段p1p2 的最小距離
db nearest(P p1, P p2, P q){
	if (p1 == p2) return p1.distTo(q);
	P h = proj(p1,p2,q);
	if(isMiddle(p1,h,p2))
		return q.distTo(h);
	return min(p1.distTo(q),p2.distTo(q));
}

// 求 線段p1p2 與 線段q1q2 的距離
db disSS(P p1, P p2, P q1, P q2){
	if(isSS(p1,p2,q1,q2)) return 0;
	return min(min(nearest(p1,p2,q1),nearest(p1,p2,q2)), min(nearest(q1,q2,p1),nearest(q1,q2,p2)));
}

db area(vector<P> ps){
	db ret = 0; rep(i,0,ps.size()) ret += ps[i].det(ps[(i+1)%ps.size()]); 
	return ret/2;
}
  
int contain(vector<P> ps, P p){ //2:inside,1:on_seg,0:outside
	int n = ps.size(), ret = 0; 
	rep(i,0,n){
		P u=ps[i],v=ps[(i+1)%n];
		if(onSeg(u,v,p)) return 1;
		if(cmp(u.y,v.y)<=0) swap(u,v);
		if(cmp(p.y,u.y) > 0 || cmp(p.y,v.y) <= 0) continue;
		ret ^= crossOp(p,u,v) > 0;
	}
	return ret*2;
}
  
vector<P> convexHull(vector<P> ps) {
	int n = ps.size(); if(n <= 1) return ps;
	sort(ps.begin(), ps.end());
	vector<P> qs(n * 2); int k = 0;
	for (int i = 0; i < n; qs[k++] = ps[i++]) 
		while (k > 1 && crossOp(qs[k - 2], qs[k - 1], ps[i]) <= 0) --k;
	for (int i = n - 2, t = k; i >= 0; qs[k++] = ps[i--])
		while (k > t && crossOp(qs[k - 2], qs[k - 1], ps[i]) <= 0) --k;
	qs.resize(k - 1);
	return qs;
}
  
vector<P> convexHullNonStrict(vector<P> ps) {
	//caution: need to unique the Ps first
	int n = ps.size(); if(n <= 1) return ps;
	sort(ps.begin(), ps.end());
	vector<P> qs(n * 2); int k = 0;
	for (int i = 0; i < n; qs[k++] = ps[i++]) 
		while (k > 1 && crossOp(qs[k - 2], qs[k - 1], ps[i]) < 0) --k;
	for (int i = n - 2, t = k; i >= 0; qs[k++] = ps[i--])
		while (k > t && crossOp(qs[k - 2], qs[k - 1], ps[i]) < 0) --k;
	qs.resize(k - 1);
	return qs;
}
  
db convexDiameter(vector<P> ps){
	int n = ps.size(); if(n <= 1) return 0;
	int is = 0, js = 0; rep(k,1,n) is = ps[k]<ps[is]?k:is, js = ps[js] < ps[k]?k:js;
	int i = is, j = js;
	db ret = ps[i].distTo(ps[j]);
	do{
		if((ps[(i+1)%n]-ps[i]).det(ps[(j+1)%n]-ps[j]) >= 0)
			(++j)%=n;
		else
			(++i)%=n;
		ret = max(ret,ps[i].distTo(ps[j]));
	}while(i!=is || j!=js);
	return ret;
}
  
vector<P> convexCut(const vector<P>&ps, P q1, P q2) {
	vector<P> qs;
	int n = ps.size();
	rep(i,0,n){
		P p1 = ps[i], p2 = ps[(i+1)%n];
		int d1 = crossOp(q1,q2,p1), d2 = crossOp(q1,q2,p2);
		if(d1 >= 0) qs.push_back(p1);
		if(d1 * d2 < 0) qs.push_back(isLL(p1,p2,q1,q2));
	}
	return qs;
}


P ld, rd, lu, ru, p[2],q[2];
void solve() {
	ld.read();
	ru.read();
	rd = P(ru.x, ld.y);
	lu = P(ld.x, ru.y);
	p[0].read(); p[1].read();
	q[0].read(); q[1].read();
	vector<P> rect{ld, rd, ru, lu};
	db ans = 0;
	for (int i = 0; i < 2; i++) for (int j = 0; j < 2; j++)
		if (p[i] == q[j]) {
			auto r = convexCut(rect, p[i], p[i] + (p[1 - i] - p[i]).rot90());
			r = convexCut(r, q[j], q[j] + (q[1 - j] - q[j]).rot90());
			ans += area(r);
		}
	if (crossOp(p[0], p[1], q[0]) == 0 && crossOp(p[0], p[1], q[1]) == 0) {
		auto r = rect;
		for (int i = 0; i < 2; i++) {
			r = convexCut(r, p[i], p[i] - (p[1 - i] - p[i]).rot90());
			r = convexCut(r, q[i], q[i] - (q[1 - i] - q[i]).rot90());
		}
		ans += area(r);
	}
	printf("%.10f\n", ans);
}

int main() {
	int tc;
	for (scanf("%d", &tc); tc; tc--) {
		solve();
	}
}

模塊3

int type(P o1,db r1,P o2,db r2){
	db d = o1.distTo(o2);
	if(cmp(d,r1+r2) == 1) return 4;
	if(cmp(d,r1+r2) == 0) return 3;
	if(cmp(d,abs(r1-r2)) == 1) return 2;
	if(cmp(d,abs(r1-r2)) == 0) return 1;
	return 0;
}
  
vector<P> isCL(P o,db r,P p1,P p2){
	if (cmp(abs((o-p1).det(p2-p1)/p1.distTo(p2)),r)>0) return {};
	db x = (p1-o).dot(p2-p1), y = (p2-p1).abs2(), d = x * x - y * ((p1-o).abs2() - r*r);
	d = max(d,(db)0.0); P m = p1 - (p2-p1)*(x/y), dr = (p2-p1)*(sqrt(d)/y);
	return {m-dr,m+dr}; //along dir: p1->p2
}
  
vector<P> isCC(P o1, db r1, P o2, db r2) { //need to check whether two circles are the same
	db d = o1.distTo(o2); 
	if (cmp(d, r1 + r2) == 1) return {};
	if (cmp(d,abs(r1-r2))==-1) return {};
	d = min(d, r1 + r2);
	db y = (r1 * r1 + d * d - r2 * r2) / (2 * d), x = sqrt(r1 * r1 - y * y);
	P dr = (o2 - o1).unit();
	P q1 = o1 + dr * y, q2 = dr.rot90() * x;
	return {q1-q2,q1+q2};//along circle 1
}
  
// extanCC, intanCC : -r2, tanCP : r2 = 0
vector<pair<P, P>> tanCC(P o1, db r1, P o2, db r2) {
	P d = o2 - o1;
	db dr = r1 - r2, d2 = d.abs2(), h2 = d2 - dr * dr;
	if (sign(d2) == 0|| sign(h2) < 0) return {};
	h2 = max((db)0.0, h2);
	vector<pair<P, P>> ret;
	for (db sign : {-1, 1}) {
		P v = (d * dr + d.rot90() * sqrt(h2) * sign) / d2;
		ret.push_back({o1 + v * r1, o2 + v * r2});
	}
	if (sign(h2) == 0) ret.pop_back();
	return ret;
}

db rad(P p1,P p2){
	return atan2l(p1.det(p2),p1.dot(p2));
}

db areaCT(db r, P p1, P p2){
	vector<P> is = isCL(P(0,0),r,p1,p2);
	if(is.empty()) return r*r*rad(p1,p2)/2;
	bool b1 = cmp(p1.abs2(),r*r) == 1, b2 = cmp(p2.abs2(), r*r) == 1;
	if(b1 && b2){
		P md=(is[0]+is[1])/2;
		if(sign((p1-md).dot(p2-md)) <= 0) 
			return r*r*(rad(p1,is[0]) + rad(is[1],p2))/2 + is[0].det(is[1])/2;
		else return r*r*rad(p1,p2)/2;
	}
	if(b1) return (r*r*rad(p1,is[0]) + is[0].det(p2))/2;
	if(b2) return (p1.det(is[1]) + r*r*rad(is[1],p2))/2;
	return p1.det(p2)/2;
}

P inCenter(P A, P B, P C) {
	double a = (B - C).abs(), b = (C - A).abs(), c = (A - B).abs();
	return (A * a + B * b + C * c) / (a + b + c);
}
 
P circumCenter(P a, P b, P c) { 
	P bb = b - a, cc = c - a;
	double db = bb.abs2(), dc = cc.abs2(), d = 2 * bb.det(cc);
	return a - P(bb.y * dc - cc.y * db, cc.x * db - bb.x * dc) / d;
}
 
P othroCenter(P a, P b, P c) { 
	P ba = b - a, ca = c - a, bc = b - c;
	double Y = ba.y * ca.y * bc.y,
	A = ca.x * ba.y - ba.x * ca.y,
	x0 = (Y + ca.x * ba.y * b.x - ba.x * ca.y * c.x) / A,
	y0 = -ba.x * (x0 - c.x) / ba.y + ca.y;
	return {x0, y0};
}

pair<P,db> min_circle(vector<P> ps){
    random_shuffle(ps.begin(), ps.end());
    int n = ps.size();
    P o = ps[0]; db r = 0;
    rep(i,1,n) if(o.distTo(ps[i]) > r + EPS){
        o = ps[i], r = 0;
        rep(j,0,i) if(o.distTo(ps[j]) > r + EPS){
            o = (ps[i] + ps[j]) / 2; r = o.distTo(ps[i]);
            rep(k,0,j) if(o.distTo(ps[k]) > r + EPS){
                 o = circumCenter(ps[i],ps[j],ps[k]);
                 r = o.distTo(ps[i]);
            }
        }
    }
    return {o,r};
}

大數

struct Bigint {
    // representations and structures
    string a; // to store the digits
    int sign; // sign = -1 for negative numbers, sign = 1 otherwise
    // constructors
    Bigint() {} // default constructor
    Bigint( string b ) { (*this) = b; } // constructor for string
    // some helpful methods
    int size() { // returns number of digits
        return a.size();
    }
    Bigint inverseSign() { // changes the sign
        sign *= -1;
        return (*this);
    }
    Bigint normalize( int newSign ) { // removes leading 0, fixes sign
        for( int i = a.size() - 1; i > 0 && a[i] == '0'; i-- )
            a.erase(a.begin() + i);
        sign = ( a.size() == 1 && a[0] == '0' ) ? 1 : newSign;
        return (*this);
    }
    // assignment operator
    void operator = ( string b ) { // assigns a string to Bigint
        a = b[0] == '-' ? b.substr(1) : b;
        reverse( a.begin(), a.end() );
        this->normalize( b[0] == '-' ? -1 : 1 );
    }
    // conditional operators
    bool operator < ( const Bigint &b ) const { // less than operator
        if( sign != b.sign ) return sign < b.sign;
        if( a.size() != b.a.size() )
            return sign == 1 ? a.size() < b.a.size() : a.size() > b.a.size();
        for( int i = a.size() - 1; i >= 0; i-- ) 
            if( a[i] != b.a[i] )
                return sign == 1 ? a[i] < b.a[i] : a[i] > b.a[i];
        return false;
    }
    bool operator == ( const Bigint &b ) const { // operator for equality
        return a == b.a && sign == b.sign;
    }
    // mathematical operators
    Bigint operator + ( Bigint b ) { // addition operator overloading
        if( sign != b.sign ) return (*this) - b.inverseSign();
        Bigint c;
        for(int i = 0, carry = 0; i<a.size() || i<b.size() || carry; i++ ) {
            carry+=(i<a.size() ? a[i]-48 : 0)+(i<b.a.size() ? b.a[i]-48 : 0);
            c.a += (carry % 10 + 48);
            carry /= 10;
        }
        return c.normalize(sign);
    }
    Bigint operator - ( Bigint b ) { // subtraction operator overloading
        if( sign != b.sign ) return (*this) + b.inverseSign();
        int s = sign; sign = b.sign = 1;
        if( (*this) < b ) return ((b - (*this)).inverseSign()).normalize(-s);
        Bigint c;
        for( int i = 0, borrow = 0; i < a.size(); i++ ) {
            borrow = a[i] - borrow - (i < b.size() ? b.a[i] : 48);
            c.a += borrow >= 0 ? borrow + 48 : borrow + 58;
            borrow = borrow >= 0 ? 0 : 1;
        }  
        return c.normalize(s);
    }
    Bigint operator * ( Bigint b ) { // multiplication operator overloading
        Bigint c("0");
        for( int i = 0, k = a[i] - 48; i < a.size(); i++, k = a[i] - 48 ) {
            while(k--) c = c + b; // ith digit is k, so, we add k times
            b.a.insert(b.a.begin(), '0'); // multiplied by 10
        }
            return c.normalize(sign * b.sign);
    }
    Bigint operator / ( Bigint b ) { // division operator overloading
        if( b.size() == 1 && b.a[0] == '0' ) b.a[0] /= ( b.a[0] - 48 );
            Bigint c("0"), d;
        for( int j = 0; j < a.size(); j++ ) d.a += "0";
        int dSign = sign * b.sign; b.sign = 1;
        for( int i = a.size() - 1; i >= 0; i-- ) {
            c.a.insert( c.a.begin(), '0');
            c = c + a.substr( i, 1 );
            while( !( c < b ) ) c = c - b, d.a[i]++;
        }
        return d.normalize(dSign);
    }
    Bigint operator % ( Bigint b ) { // modulo operator overloading
        if( b.size() == 1 && b.a[0] == '0' ) b.a[0] /= ( b.a[0] - 48 );
        Bigint c("0");
        b.sign = 1;
        for( int i = a.size() - 1; i >= 0; i-- ) {
            c.a.insert( c.a.begin(), '0');
            c = c + a.substr( i, 1 );
            while( !( c < b ) ) c = c - b;
        }
        return c.normalize(sign);
    }
    // output method
    void print() {
        if( sign == -1 ) putchar('-');
        for( int i = a.size() - 1; i >= 0; i-- ) putchar(a[i]);
    }
};

雜

內建函數

`__builtin_ctz( )`

返回輸入數二進制表示從最低位開始(右起)的連續的0的個數；如果傳入0則行為未定義。

`__builtin_parity( )`

返回括號內數的二進制表示形式中1的個數的奇偶性（偶：0，奇：1）。

`__builtin_popcount( )`

返回括號內數的二進制表示形式中1的個數。

過去犯錯

我手比較快，可以拼下手速，如果卡題了，可以拼一下題數，反正實力在那里了，好幾場VP省賽都金銀，不會差的
注意細心讀題，不要忽略任何一個條件
結果錯了，輸出運算時候的中間值檢驗；
代碼在本地編譯錯誤，找不出，一段一段刪除代碼然后編譯看是哪部分有問題（ctrl+z可以返回上一步，要注意你的編譯器有沒有這個功能）
代碼過了樣例，交上去錯了，檢查數組有沒有開小，有沒有用long long甚至要高精度計算（注意題目給出的數據范圍），或者是運算的時候超出了int的范圍使得結果不正確等。
你在主函數的數組開的太大了，導致你程序無法輸入等問題，開成全局變量就可以解決
盡量少用指針，算法競賽的題目不應該出現復雜的指針，隊列，鏈表，樹和圖的存儲可以用數組實現（鏈式前向星）或鄰接矩陣。學習算法競賽中使用的存圖方式
你數組開小了，題目的數據范圍是1e5，那你就要開到1e5+10,
數組開小，或者開大可能會導致WA,RE,MLE,TLE等，具體問題具體分析
代碼盡量不要讓別人幫你看，前期還是提升個人能力為主
一些C++20編譯器你RE了，他會爆TLE，坑了我好幾次了，嗚嗚，換C++17,C++11，C++14報錯正常，具體看評測機，再遇到再補充吧
set判重速度要慢一倍（特定情況下），用了數組才過。Acwing的周賽連通塊
異或運算要加括號，運算優先級的問題
set可以lower_bound

例如int op=*x.lower_bound(x1);
x是set<int>::iterator it; it=se.lower_bound(9);

map<int, string>::iterator p1, p2; p1 = mp.lower_bound(3); /* p1->second == 3 */
左移運算符 a = 1ll << 60; 才是對的
multiset 刪除 s.erase(s.find(a));

posted on 2024-10-12 17:33 nannandbk 閱讀(209) 評論(0) 收藏舉報

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Math & DP & Data structure & Graph & Geometry & Game

Head

快讀1

快讀2

快讀3

模板1

模板2

模板3

Math

快速冪

快速乘

除法取整

龜速乘

數論分塊

枚舉子集，超集

矩陣乘法

矩陣運算及求行列式

線性篩

埃式篩

區間篩

質因數分解

歐拉函數

約數

最大公約數

求約數

擴展歐幾里得1

擴展歐幾里得2

線性同余方程

【模板】二元一次不定方程 (exgcd)

中國剩余定理

dls

ex中國剩余定理

逆元

線性

快速冪求逆元

階 \(a ^ x \equiv 1 (\bmod p)\)

原根 \(a ^ x \equiv 1 (\bmod m)\) ，當最小的 \(x\) 是 \(\varphi(m)\) 的時候 \(a\) 是 \(m\) 的原根

指數方程1 \(a ^ x \equiv b (\bmod m)\) \(m\) 為質數

指數方程2 \(a ^ x \equiv b (\bmod m)\) \(m\) 不為質數

排列組合

加法 & 乘法原理

加法原理

乘法原理

組合數

\(O(N^2)\)

O(N)

Lucas \(O(\log_pNp\log_p) p 為質數\)

Lucas \(O(\log_qNq\log_q)\)

莫比烏斯反演

積性函數

\(\varphi\)

\(\mu\)

杜教篩

大素數判斷 miller_rabin \(O(s \log_2 ^ 3 n)\)， \(n \leq 2 ^ {54}\)

勾股數的神奇性質

DP

01背包

完全背包

多重背包

\(O(NMS)\)

\(O(NM\log S)\)

\(O(NM)\) ver 1

\(O(NM)\) ver 2

分組背包

最長公共子序列

最長上升子序列

\(O(N^2)\)

\(O(N\log N)\)

樹上背包

\(O(N^2)\) 點數和背包大小都為\(N\)

\(O(NM^2)\)

數位DP

區間DP

Data structure

單調棧

單調隊列

并查集

普通板子

封裝DSU

帶權并查集

枚舉子集，超集

原根 \(a ^ x \equiv 1 (\bmod m)\) ，當最小的 \(x\) 是 \(\varphi(m)\) 的時候 \(a\) 是 \(m\) 的原根

大素數判斷 miller_rabin \(O(s \log_2 ^ 3 n)\)， \(n \leq 2 ^ {54}\)

樹鏈剖分點權

樹鏈剖分邊權下放

Johnson 全源最短路稀疏圖\(O(NM \log M)\) 最慢 \(O((N + M) \times (C + 1))\)，\(C\) 是環數

Prim \(O(N^2)\)，稀疏圖可以用堆優化\(O(M\log N)\)，但Kruskal更好