POJ2379---ACM Rank Table

Language:Default ACM Rank Table Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 5785   Accepted: 1475 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:  Each run is either accepted or rejected.  The problem is considered solved by the team, if one of the runs submitted for it is accepted.  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.  The total time is the sum of the time consumed for each problem solved.  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.  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.  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 3 1 2 3000 0 1 2 3100 1 2 1 4200 1 Sample Output 2 1 3

 

个人吐槽：很坑的一道题，有没有！

/*
这道题需要注意的是：
  a.一次都没有提交过的队伍
  b.已经WA了的
  c.罚时一个是分一个是秒
  d.不是按时间顺序给的
*/
#include<cstdio>
#include<cstring>
#include<string>
#include<algorithm>
#include<iostream>
using namespace std;
int N,C;
struct node1
{
    int c1,flag[1005],t[1005],numbers,times;
}a[10005];
struct node2
{
     int ci,ti,pi,ri;
}b[10005];
bool cmp1(node2 x,node2 y)
{
    return x.ti<y.ti; //先按时间顺序排列
}
bool cmp2(node1 x,node1 y)//三层排序
{
if(x.numbers!=y.numbers)//第一层：通过题数多的在前
    return x.numbers>y.numbers;
else if(x.times!=y.times)//第二层：通过题数相同：用时少的在前
    return x.times<y.times;
else//第三层：题数、时间相同时，队列编号小的在前
    return x.c1<y.c1;
}
int main()
{
   scanf("%d%d",&C,&N);
       for(int i=1;i<=C;i++)//对结构体node1数据初始化
       {
           memset(a[i].flag,0,sizeof(a[i].flag));
           memset(a[i].t,0,sizeof(a[i].t));
           a[i].numbers=a[i].times=0;
           a[i].c1=i;
       }
       for(int i=1;i<=N;i++)
        scanf("%d%d%d%d",&b[i].ci,&b[i].pi,&b[i].ti,&b[i].ri);
       sort(b+1,b+N+1,cmp1);//时间排序
       for(int i=1;i<=N;i++)
       {
          int c2=b[i].ci;
          int p2=b[i].pi;
          int t2=b[i].ti;
          int r2=b[i].ri;
          if(a[c2].flag[p2])continue;//已经AC过
          if(r2)//不是一次AC
          {
              a[c2].flag[p2]=1;//标记
              a[c2].numbers++;//通过题数+1
              a[c2].times+=t2+a[c2].t[p2];//罚时加第一次AC提交的时间
          }
          else//WA
            a[c2].t[p2]+=1200;//加二十分钟罚时
       }
         sort(a+1,a+C+1,cmp2);//三层排序
        for(int i=1;i<=C;i++)printf("%d%c",a[i].c1,i==C?'\n':' ');
        return 0;
}