luogu 模板：https://www.luogu.com.cn/problem/P3379

樹剖求 \(LCA\)
時間復雜度 \(O(2n + qlogn)\)

#include<bits/stdc++.h>
using namespace std;
using LL = long long;
struct HLD{
	vector<vector<int>> e;
	vector<int> top, dep, parent, siz, son;
	HLD(int n){
		e.resize(n + 1);
		top.resize(n + 1);
		dep.resize(n + 1);
		parent.resize(n + 1);
		siz.resize(n + 1);
		son.resize(n + 1);
	}
	void add(int u, int v){
		e[u].push_back(v);
		e[v].push_back(u);
	}
	void dfs1(int u){
		siz[u] = 1;
		dep[u] = dep[parent[u]] + 1;
		for (auto v : e[u]){
			if (v == parent[u]) continue;
			parent[v] = u;
			dfs1(v);
			siz[u] += siz[v];
			if (siz[v] > siz[son[u]]) son[u] = v;
		}
	}
	void dfs2(int u, int up){
		top[u] = up;
		if (son[u]) dfs2(son[u], up);
		for (auto v : e[u]){
			if (v == parent[u] || v == son[u]) continue;
			dfs2(v, v);
		}
	}
	int lca(int u, int v){
		while (top[u] != top[v]){
			if (dep[top[u]] > dep[top[v]]){
				u = parent[top[u]];
			}
			else{
				v = parent[top[v]];
			}
		}
		return dep[u] < dep[v] ? u : v;
	}
};
int main(){
	ios::sync_with_stdio(false);cin.tie(0);
	int n, m, s;
	cin >> n >> m >> s;
	HLD t(n);
	for (int i = 0; i < n - 1; i ++ ){
		int u, v;
		cin >> u >> v;
		t.add(u, v);
	}
	t.dfs1(s);
	t.dfs2(s, s);
	for (int i = 0; i < m; i ++ ){
		int u, v;
		cin >> u >> v;
		cout << t.lca(u, v) << "\n";
	}
	return 0;
}

倍增求 \(LCA\)
時間復雜度 預處理時間：\(O(nlogn)\)

#include <bits/stdc++.h>
using namespace std;
const int N = 5e5 + 10;
int n, m, root, d[N], p[N][30], lg[N];
vector <int> g[N];
void dfs(int u, int fa){
	p[u][0] = fa;
	d[u] = d[fa] + 1;
	for (int i = 1; i <= lg[d[u]]; i ++ )
		p[u][i] = p[p[u][i - 1]][i - 1];
		// u 的 2^i 的祖先等于 u 的 2^(i-1) 的祖先的 2^(i-1) 的祖先
	for (auto v : g[u])
		if (v != fa)
			dfs(v, u);
}
int lca(int x, int y){
	if(d[x] < d[y]) swap(x, y);
	while (d[x] > d[y])
		x = p[x][lg[d[x] - d[y]] - 1];
	if (x == y) return x;
	for (int k = lg[d[x]] - 1; k >= 0; k -- )
		if (p[x][k] != p[y][k]){
			x = p[x][k];
			y = p[y][k];
		}
	return p[x][0];
}
int main(){
	ios::sync_with_stdio(false);cin.tie(0);
	cin >> n >> m >> root;
	for (int i = 1; i < n; i ++ ){
		int u, v;
		cin >> u >> v;
		g[u].push_back(v);
		g[v].push_back(u);
	}
	for (int i = 1; i <= n; i ++ )	//預處理 log
		lg[i] = lg[i - 1] + (1 << lg[i - 1] == i);
	dfs(root, 0);	//找到每個點的祖先 
	for (int i = 1; i <= m; i ++ ){
		int x, y;
		cin >> x >> y;
		cout << lca(x, y) << "\n";
	}
	return 0;
}