poj 3660 Cow Contest

Cow Contest

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 8987		Accepted: 5046

Description

N (1 ≤ N ≤ 100) cows, conveniently numbered 1..N, are participating in a programming contest. As we all know, some cows code better than others. Each cow has a certain constant skill rating that is unique among the competitors.

The contest is conducted in several head-to-head rounds, each between two cows. If cow A has a greater skill level than cow B (1 ≤ A ≤ N; 1 ≤ B ≤ N; A ≠ B), then cow A will always beat cow B.

Farmer John is trying to rank the cows by skill level. Given a list the results of M (1 ≤ M ≤ 4,500) two-cow rounds, determine the number of cows whose ranks can be precisely determined from the results. It is guaranteed that the results of the rounds will not be contradictory.

Input

* Line 1: Two space-separated integers: N and M
* Lines 2..M+1: Each line contains two space-separated integers that describe the competitors and results (the first integer, A, is the winner) of a single round of competition: A and B

Output

* Line 1: A single integer representing the number of cows whose ranks can be determined
　

Sample Input

5 5
4 3
4 2
3 2
1 2
2 5

Sample Output

2

可以说得上是最短路吧，反正我会是用弗洛伊德暴力水过的，代码很简单，思路清晰就好。

题意：牛之间有绝对的强弱，给出一些胜负关系，问有多少头牛可以确定其绝对排名。

附上代码：

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
int main()
{
    int n,m,i,j,k;
    while(~scanf("%d%d",&n,&m))
    {
        int a,b,map[105][105],t;
        memset(map,0,sizeof(map));
        for(i=1; i<=m; i++)
        {
            scanf("%d%d",&a,&b);
            map[a][b]=1;
        }
        for(i=1; i<=n; i++)
            for(j=1; j<=n; j++)
                for(k=1; k<=n; k++)
                    if(map[j][i]&&map[i][k])  //暴力弗洛伊德，所有的关系连成图
                        map[j][k]=1;
        int ans=0;
        for(i=1; i<=n; i++)
        {
            int t=0;
            for(j=1; j<=n; j++)
                t+=map[i][j]+map[j][i];
            if(t==n-1)
                ans++;
        }
        printf("%d\n",ans);
    }
    return 0;
}