[POJ] 2379 -> ACM Rank Table

ACM Rank Table

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 4523 Accepted: 1188

Description

ACM contests, like the one you are participating in, are hosted by the special software. That software, among other functions, preforms a job of accepting and evaluating teams' solutions (runs), and displaying results in a rank table. The scoring rules are as follows:

1. Each run is either accepted or rejected.
   
2. The problem is considered solved by the team, if one of the runs submitted for it is accepted.
   
3. The time consumed for a solved problem is the time elapsed from the beginning of the contest to the submission of the first accepted run for this problem (in minutes) plus 20 minutes for every other run for this problem before the accepted one. For an unsolved problem consumed time is not computed.
   
4. The total time is the sum of the time consumed for each problem solved.
   
5. Teams are ranked according to the number of solved problems. Teams that solve the same number of problems are ranked by the least total time.
   
6. While the time shown is in minutes, the actual time is measured to the precision of 1 second, and the the seconds are taken into account when ranking teams.
   
7. Teams with equal rank according to the above rules must be sorted by increasing team number.

Your task is, given the list of N runs with submission time and result of each run, compute the rank table for C teams.

Input

Input contains integer numbers C N, followed by N quartets of integes ci pi ti ri, where ci -- team number, pi -- problem number, ti -- submission time in seconds, ri -- 1, if the run was accepted, 0 otherwise.
1 ≤ C, N ≤ 1000, 1 ≤ ci ≤ C, 1 ≤ pi ≤ 20, 1 ≤ ti ≤ 36000.

Output

Output must contain C integers -- team numbers sorted by rank.

Sample Input

3 31 2 3000 01 2 3100 12 1 4200 1

Sample Output

2 1 3

解题思路：

可能出现以下几个问题：

1.ac了的题目重复提交

2.有可能不按时间顺序排列

3.没k题过的队伍也要算上

4.提交时间和罚时单位一个是秒一个是分钟

Code:

#include <iostream>#include <vector>#include <algorithm>using namespace std;typedef struct Team{int ci;int count;int time;int wrong[21];Team() : count(0), time(0) { memset(wrong, 0, 21 * sizeof(int)); }} Team;typedef struct Submission{int ci, pi, ti;bool ri;Submission(int c, int p, int t, bool r) : ci(c), pi(p), ti(t), ri(r) {}}Submission;bool cmp_subs(Submission& sub1, Submission sub2){return sub1.ti < sub2.ti;}bool cmp_teams(Team& t1, Team& t2){if (t1.count != t2.count)return t1.count > t2.count;else if (t1.time != t2.time)return t1.time < t2.time;else return t1.ci < t2.ci;}int main(){int C, N, ci, pi, ti;bool ri;cin >> C >> N;Team* teams = new Team[C+1];vector<Submission> subs;for (int i = 1; i <= C + 1; ++i)teams[i].ci = i;for (int i = 0; i < N; i++){cin >> ci >> pi >> ti >> ri;Submission sub(ci, pi, ti, ri);subs.push_back(sub);}sort(subs.begin(), subs.end(), cmp_subs);for (vector<Submission>::iterator it = subs.begin(); it != subs.end(); ++it){if (teams[it->ci].wrong[it->pi] != -1) { //如果没有提交（有些白痴队伍ac了还要重复提交）if (it->ri == true){teams[it->ci].time += teams[it->ci].wrong[it->pi] * 20 * 60 + it->ti;++teams[it->ci].count;teams[it->ci].wrong[it->pi] = -1;} else ++teams[it->ci].wrong[it->pi];}}sort(teams + 1, teams + C + 1, cmp_teams);for (int i = 1; i < C; i++)cout<<teams[i].ci<<" ";cout << teams[C].ci << endl;return 0;}