Cow Contest POJ - 3660 （floyd 传递闭包）

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

题意：n个人，m个关系，每个关系（a b）告诉你,a比b水平高，问你最后有多少人水平是可以确定的。
思路:刚开始一开这题意就莽拓扑排序，后来发现不对，（想着如果入队多个，说明无序，入队一个就有序，简直睿智）
其实是用floyd计算传递闭包，最后看对每个人是否关系都确定了。

（因为我图中只记录了，谁比谁等级高，所以闭包出来，也只有该点是否比其他点等级高，例如 maps【a】【b】，显然缺少一种别点比他高的情况，其实就是maps【b】【a】）

 #include<cstdio>
 #include<iostream>

 int maps[][];

 int n,m;
 int main()
 {
     scanf("%d%d",&n,&m);
     for(int j=; j<=m; j++)
     {
         int u,v;
         scanf("%d%d",&u,&v);
         maps[u][v] = ;
     }
     for(int k=; k<=n; k++)
     {
         for(int i=; i<=n; i++)
         {
             for(int j=; j<=n; j++)
             {
                 maps[i][j] |= maps[i][k] && maps[k][j];
             }
         }
     }
     int num=n;
 //    for(int i=1;i<=n;i++)
 //    {
 //        for(int j=1;j<=n;j++)
 //        {
 //            printf("%d  ",maps[i][j]);
 //        }
 //        puts("");
 //    }
     for(int i=; i<=n; i++)
     {
         for(int j=; j<=n; j++)
         {
             if(j == i)continue;
             if(!maps[i][j] && !maps[j][i])
             {
                 num--;
                 break;
             }
         }
     }
     printf("%d\n",num);
 }