Floyd最短路——POJ 3360 Cow Contest

Cow Contest

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 13133		Accepted: 7312

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

啊啊啊啊！最短路啊最短路啊最短路啊啊啊！~~~本题完全可以用Floyd的中转思想确定cowA与cowB的关系，是A强于B，A弱于B，亦或是AB没有关系。若A强于B，B强于C，那么A一定强于C！Floyd三层循环每次更新不同牛之间的关系。距离用强弱关系表示即可（这一点当时没想到所以并不知道和最短路有什么关系。。。醉醉的。。。）， 值为1表示A强于B，为-1表示A弱于B，为0表示AB无关系，若比A强的牛与比A弱牛之和为N-1,那么A的位置也就是名次一定是确定的，因此最后只需检查更新后的表中有几头这样的牛即可！真是遇题不能拖啊，感谢学长大人的开导哇，让我对Floyd有更深的理解/抱拳 附代码：

#include
int main()
{
    int n,m;
    while(scanf("%d%d",&n,&m)!=EOF)
    {
        int a[110][110];
        for(int i=1;i<=n;i++)
            for(int j=1;j<=n;j++)
                  a[i][j]=0;
        for(int i=1;i<=m;i++)
        {
            int x,y;
            scanf("%d%d",&x,&y);
            a[x][y]=1;
            a[y][x]=-1;
        }
        for(int k=1;k<=n;k++)
          for(int i=1; i<=n;i++)
            for(int j=1;j<=n;j++)
               if(a[i][k]==1&&a[k][j]==1)
               {
                a[i][j]=1;
                a[j][i]=-1;
               }
        int sum=0;
        for(int i=1;i<=n;i++)
        {
            int q=0;
            for(int j=1;j<=n;j++)
            {
                if(i!=j&&(a[i][j]==-1||a[i][j]==1))
                    q++;
            }
            if(q==n-1)
                sum++;
        }
        printf("%d\n",sum);
    }
    return 0;
}