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.

* 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

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

 1 #include <iostream>
 2 #include <cstdio>
 3 using    namespace    std;
 4
 5 const    int    SIZE = 105;
 6 int    N,M;
 7 int    IN[SIZE],OUT[SIZE];
 8 bool    G[SIZE][SIZE];
 9
10 int    main(void)
11 {
12     int    from,to;
13     int    ans;
14
15     while(scanf("%d%d",&N,&M) != EOF)
16     {
17         fill(&G[0][0],&G[N][N],false);
18         for(int i = 0;i <= N;i ++)
19             IN[i] = OUT[i] = 0;
20
21         for(int i = 0;i < M;i ++)
22         {
23             scanf("%d%d",&from,&to);
24             G[from][to] = true;
25         }
26
27         for(int k = 1;k <= N;k ++)
28             for(int i = 1;i <= N;i ++)
29                 for(int j = 1;j <= N;j ++)
30                     G[i][j] = G[i][j] || G[i][k] && G[k][j];
31         for(int i = 1;i <= N;i ++)
32             for(int j = 1;j <= N;j ++)
33                 if(G[i][j])
34                 {
35                     IN[j] ++;
36                     OUT[i] ++;
37                 }
38
39         ans = 0;
40         for(int i = 1;i <= N;i ++)
41             if(IN[i] + OUT[i] == N - 1)
42                 ans ++;
43         printf("%d\n",ans);
44     }
45
46     return    0;
47 }