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數組的強大作用

Description

As a contestant, you must be familiar with the rules of ACM-ICPC. Teams are ranked according to the most problems solved. Teams who solve the same number of problems are ranked by least total time. The total time is the sum of the time consumed for each problem solved. The time consumed for a solved problem is the time elapsed from the beginning of the contest to the submittal of the accepted run plus 20 penalty minutes for every rejected run for that problem regardless of submittal time. There is no time consumed for a problem that is not solved. Teams who solve the same number of problems and the same total time are ranked by their names in alphabetical order.

Any submit of a problem by those teams who has solved this problem before would be ignored!

Caspar (a coach of SCPC) wants to know when the rank of his students’ team changed. He makes a curve to show the rank’s changes of his students’ team. Now, give you the information of every submission. Your problem is to calculate the changing points of a giving team.

Input

There is ONLY ONE TESTCASE.

The first line is an integer N (1<=N<=200) which indicates the number of submissions, and the number of teams is less than 20.

Each of the following N lines has four operations A (indicating the Name of the team who submitted this code, no more than 20 characters), B(an integer indicating the submitted time), C (indicating the ID of problem, only one uppercase character), D (indicating the result of this run, ‘Y’ means accepted and ‘N’ means not accepted).

Then an integer M followed in one line.

Each of the following M lines has only one string indicating the team name which Caspar wants to check its curve.

We promise that there is at most one submit in the same time and each of those teams Caspar want to check has at least one submit in this competition.

Output

For each team that Caspar wants to check its curve. You should output the changing points of the team in several lines. Each line has two number T, R indicating the time when the rank has been changed and rank. You should output as this format “time:T rank:R”. There is a blank line followed.

After that, you should output the final rank of all teams.

Sample Input

16
TEAM1 1 A Y
TEAM2 2 A N
TEAM2 3 A Y
Dream 4 A Y
TEAM3 5 A Y
TEAM3 6 B Y
TEAM4 7 A Y
TEAM4 8 B Y
TEAM4 9 C Y
Dream 10 B Y
Dream 11 C Y
TEAM3 12 C Y
TEAM1 13 B Y
TEAM1 14 C Y
TEAM1 15 D Y
Dream 16 D Y
2
Dream
TEAM1

Sample Output

time:4 rank:2
time:6 rank:3
time:8 rank:4
time:10 rank:3
time:11 rank:2
time:12 rank:3
time:15 rank:4
time:16 rank:1

time:1 rank:1
time:6 rank:2
time:8 rank:3
time:10 rank:4
time:15 rank:1
time:16 rank:2

Dream
TEAM1
TEAM3
TEAM4
TEAM2

#include "iostream"
#include "cstring"
#include "algorithm"
#include "string"
#include "map"
using namespace std;
#define sz 21
#define ca 201
int tm[sz][ca], rk[sz][ca], wg[sz][ca];
bool AC[sz][ca];
struct submit
{
    string na;
    int ti;
    char pr, sb;
}s[ca];
struct team
{
    string na;
    int an, at, ra;
}t[sz], tt[sz];
bool check(team a, team b)
{
    if(a.an > b.an) return true;
    if(a.an==b.an && a.at<b.at) return true;
    if(a.an==b.an && a.at==b.at && a.na<b.na) return true;
    return false;
}
int main()
{
    int n, k1 = 0, k2 = 0, flag1, flag2, df[sz], who, whi, nowtime;
    string ne;
    map<string, int> nm; //name
    map<char, int> pm; //probelm
    for(int i=0; i<21; i++)
    {
        t[i].an = 0; t[i].at = 0;
        t[i].ra = 0; df[i] = 0;
    }
    memset(tm, 0, sizeof(tm)); memset(rk, 0, sizeof(rk));
    memset(wg, 0, sizeof(wg)); memset(AC, false, sizeof(AC));
    cin>>n;
    for(int i=0; i<n; i++)
    {
        flag1=1; flag2=1;
        cin>>s[i].na>>s[i].ti>>s[i].pr>>s[i].sb;
        nowtime = s[i].ti;
        for(int j=0; j<i; j++)
        {
            if(s[i].na==s[j].na) flag1=0;
            if(s[i].pr==s[j].pr) flag2=0;
        }
        if(flag1) 
        {
            t[k1].na = s[i].na; //存入姓名
            nm[s[i].na]=k1; k1++;
        }
        if(flag2)
        {
            pm[s[i].pr]=k2;
            k2++;
        }
        who = nm[s[i].na]; whi = pm[s[i].pr];
        if(s[i].sb=='N') wg[who][whi]++;
        else if(AC[who][whi]==false)
        {
            AC[who][whi]=true;
            t[who].an++; 
            t[who].at+=(nowtime+20*wg[who][whi]);
        }
        for(int i=0; i<k1; i++) tt[i] = t[i];
        sort(tt, tt+k1, check);
        for(int j=0; j<k1; j++)
        {
            who = nm[tt[j].na]; 
            if(t[who].ra!=(j+1)) //j+1是現在的排名
            {
                t[who].ra = j+1; df[who]++;
                tm[who][df[who]]=nowtime; //記錄提交時的時間
                rk[who][df[who]]=j+1;
            }
        }
    }
    cin>>n;
    while(n--)
    {
        cin>>ne;
        for(int i=1; i<=df[nm[ne]]; i++) cout<<"time:"<<tm[nm[ne]][i]<<" rank:"<<rk[nm[ne]][i]<<endl;
        cout<<endl;
    }
    for(int i=0; i<k1; i++) cout<<tt[i].na<<endl;
    return 0;
}

posted on 2011-11-21 21:20 More study needed. 閱讀(432) 評論(0) 收藏舉報

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書山有徑勤為路>>>>>>>>

<<<<<<<<學海無涯苦作舟!

數組的強大作用

Sample Input

Sample Output

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