zoukankan      html  css  js  c++  java
  • HNUST-1148 ACM ranking rules(简单模拟)

    1148: ACM ranking rules

    时间限制: 1 Sec  内存限制: 128 MB
    提交: 16  解决: 12
    [提交][状态][讨论版]

    题目描述

    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.

    输入

    Input contains integer numbers C N, followed by N quartets of integers 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.

    输出

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

    样例输入

    3 3
    1 2 3000 0
    1 2 3100 1
    2 1 4200 1
    

    样例输出

    2 1 3
    

    提示

    用结构体数组存储各队信息

    #include<algorithm>
    #include<iostream>
    #include<cstdio>
    #include<map>
    using namespace std;
     
    const int N = 1000 + 5;
     
    int c, n;
    struct node{
        short _time = 0, solve = 0, num;
        map<int, int> state;
    }Node[N];
     
    int main(){
        scanf("%d %d", &c, &n);
        for(int i = 1; i <= c; i++) Node[i].num = i;
     
        int ci, pi, ti, ri;
        for(int i = 1; i <= n; i++){
            scanf("%d %d %d %d", &ci, &pi, &ti, &ri);
            auto & mp = Node[ci].state;
            if(mp[pi] < 0) continue;
            if(ri == 0 ) mp[pi]++;
            if(ri == 1) {
                Node[ci]._time += (mp[pi] * 20 * 60 + ti);
                mp[pi] = -1;
                Node[ci].solve++;
            }
        }
        auto cmp = [](const node &x, const node &y) -> bool{
            if(x.solve != y.solve) return x.solve > y.solve;
            return x._time < y._time;
        };
        stable_sort(Node + 1, Node + n + 1, cmp);
        for(int i = 1; i <= n; i++){
            printf("%d%c", Node[i].num, (i == n)?'
    ':' ');
        }
    }
  • 相关阅读:
    Unity 关于特效和UI显示的优先级问题
    使用Frida神器轻松实现hook C/C++方法
    理解 Android Binder 机制(三):Java层
    理解 Android Binder 机制(二):C++层
    理解 Android Binder 机制(一):驱动篇
    Android Hook Instrumentation
    Cocos Creator 中根据uuid快速定位资源
    android 通用混淆配置
    vToRay + bbr 加速
    SpringBoot项目单元测试
  • 原文地址:https://www.cnblogs.com/Pretty9/p/7784912.html
Copyright © 2011-2022 走看看