POJ 3660 Cow Contest (闭包传递)

Cow Contest

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 7690		Accepted: 4288

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

刚才学了下闭包传递，简单来说就是用Floyd来求两点是否连通。这题里，如果一点的入度加上出度等于N-1，那么就可确定这点的等级，因为除了它自己之外的所有节点要么在它之前要么在它之后，不管它前面的节点后后面的节点怎么排序，都不会影响到它。

 #include <iostream>
 #include <cstdio>
 using    namespace    std;

 const    int    SIZE = ;
 int    N,M;
 int    IN[SIZE],OUT[SIZE];
 bool    G[SIZE][SIZE];

 int    main(void)
 {
     int    from,to;
     int    ans;

     while(scanf("%d%d",&N,&M) != EOF)
     {
         fill(&G[][],&G[N][N],false);
         for(int i = ;i <= N;i ++)
             IN[i] = OUT[i] = ;

         for(int i = ;i < M;i ++)
         {
             scanf("%d%d",&from,&to);
             G[from][to] = true;
         }

         for(int k = ;k <= N;k ++)
             for(int i = ;i <= N;i ++)
                 for(int j = ;j <= N;j ++)
                     G[i][j] = G[i][j] || G[i][k] && G[k][j];
         for(int i = ;i <= N;i ++)
             for(int j = ;j <= N;j ++)
                 if(G[i][j])
                 {
                     IN[j] ++;
                     OUT[i] ++;
                 }

         ans = ;
         for(int i = ;i <= N;i ++)
             if(IN[i] + OUT[i] == N - )
                 ans ++;
         printf("%d\n",ans);
     }

     return    ;
 }