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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

2

输入a,b，表示a能打败b，给出多组这种数据，求能够确定能力排行的牛有多少个。

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            #define INF 0x3f3f3f3f#define MAXN 5005#define mod 1000000007using namespace std;int n,m;int map[MAXN][MAXN];void Floyd(){ int i,j,k; for(k=1; k<=n; k++) for(i=1; i<=n; i++) for(j=1; j<=n; j++) if(map[i][j]>map[i][k]+map[k][j]) map[i][j]=map[i][k]+map[k][j];}int main(){ scanf("%d%d",&n,&m); for(int i=1; i<=n; i++) for(int j=1; j<=n; j++) map[i][j]=(i==j?0:INF); //注意自身和自身距离为0。 int a,b; while(m--) { scanf("%d%d",&a,&b); map[a][b]=1; } Floyd(); int ans=0; for(int i=1;i<=n;++i) { int tmp=0; for(int j=1;j<=n;++j) { if(i!=j) { if(map[i][j]
           
          
         
        
       
      
     
    
   

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