POJ-2379 ACM Rank Table 模拟题

ACM Rank Table

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 2979		Accepted: 764

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

Sample Output

2 1 3

　　模拟题，注意数据中的时间是以秒为计数单位，罚时是20分钟，所以要乘以60秒钟。还有一点就是可能会出现多次提交同一个题目，就算那道题已经AC了，无聊啊，而且还有可能先告诉你
某队在后面的时间错误提交，而后面的数据有告诉你前面他已经AC了这道题目。
　　

#include <stdio.h>
#include <string.h>
#include <stdlib.h>

struct T
{
	int team, pnum, time;
	int pwrong[21];
}t[1005];

struct D
{
	int c, p, ti,r; 
}d[1005];

int C, N;

int cmp( const void *a, const void *b )
{
	struct T *t1= ( struct T *)a, *t2= ( struct T * )b; 
	if( t1-> pnum!= t2-> pnum )
	{
		return t2-> pnum- t1-> pnum;
	}
	else if( t1-> time!= t2-> time )
	{
		return t1-> time- t2-> time;
	}
	else
	{
		return t1-> team- t2-> team;
	}
}

int cmp2( const void *a, const void *b )
{
	struct D *d1= ( struct D * )a, *d2= ( struct D * )b;
	return d1-> ti- d2-> ti;
}

int main(  )
{
	while( scanf( "%d %d", &C, &N )!= EOF )
	{
		memset( t, 0, sizeof( t[0] )* C );
		for( int i= 1; i<= C; ++i )
		{
			t[i]. team= i;
		}
		for( int i= 0; i< N; ++i )
		{
			scanf( "%d %d %d %d", &d[i]. c, &d[i]. p, &d[i]. ti, &d[i]. r );
		}
		qsort( d, N, sizeof( d[0] ), cmp2 );
		for( int i= 0; i< N; ++i )
		{
			int c= d[i]. c, p= d[i]. p, ti= d[i]. ti, r= d[i]. r;
			if( r== 1 )
			{	
				if( t[c]. pwrong[p]!= -1 )
				{
					t[c]. time+= ( ti+ t[c]. pwrong[p]* 20* 60 );
					t[c]. pnum++;
				}
				t[c]. pwrong[p]= -1;
			}
			else
			{
				if( t[c]. pwrong[p]!= -1 )
				{
					t[c]. pwrong[p]++;
				}
			}
		}
		qsort( t+ 1, C, sizeof( t[0] ), cmp );
		for( int i= 1; i<= C; ++i )
		{
			printf( i== 1? "%d": " %d", t[i]. team );
		}
		puts( "" );
	}
	return 0;
}

代码如下：