（poj 3660） Cow Contest （floyd算法+传递闭包）

题目链接：http://poj.org/problem?id=3660

Description

N ( ≤ N ≤ ) cows, conveniently numbered ..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 ( ≤ A ≤ N;  ≤ 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 ( ≤ M ≤ ,) 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 : Two space-separated integers: N and M
* Lines ..M+: 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 : A single integer representing the number of cows whose ranks can be determined
　

Sample Input

Sample Output

Source
USACO  January Silver

题目大意：有N头牛，每头牛都有个特技。有M场比赛，比赛形式是A B 意味着牛A一定能赢牛B，问在M场比赛后，有几头牛的名次是确定的（不互相矛盾）？

方法：这是最短路练习，但是以最短路的形式无法求出来，看了一下题解发现是最短路的floyd算法写的，这个算法的时间复杂度为O(n3),自己百度了一下这个算法，再求每个点的出度入度的和，如果等于n（算上自己），就能确定

 #include<stdio.h>
 #include<cstring>
 #include<cstdlib>
 #include<algorithm>
 #include<math.h>
 #include<queue>
 using namespace std;
 #define INF 0x3f3f3f3f
 #define ll long long
 #define met(a,b) memset(a,b,sizeof(a))
 #define N 500
 int a[N][N],G[N][N];
 int main()
 {
     int n,m,e,v;
     while(scanf("%d %d",&n,&m)!=EOF)
     {
         for(int i=;i<=n;i++)
         {
             for(int j=;j<=n;j++)
                 a[i][j]=i==j?:;
         }
         for(int i=;i<m;i++)
         {
             scanf("%d %d",&e,&v);
               a[e][v]=;
         }
         ///floyd算法 五行代码 三层for循环   时间复杂度为O（n3）
         for(int k=;k<=n;k++)
         {
             for(int i=;i<=n;i++)
             {
                 for(int j=;j<=n;j++)
                 {
                     if(a[i][j]||(a[i][k] && a[k][j]))
                         a[i][j]=;
                 }
             }
         }
         ///求每个点的出度入度之和
         int ans=,sum=;
         for(int i=;i<=n;i++)
         {
             sum=;
             for(int j=;j<=n;j++)
             {
                 if(a[i][j] || a[j][i])
                     sum++;
             }
             if(sum==n)
                 ans++;
         }
         printf("%d\n",ans);
     }
     return ;
 }
 /*
 5 5
 4 3
 4 2
 3 2
 1 2
 2 5
 */