ayit第十六周周赛题 a题

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

 

#include
#include 
int g[110][120],n,m,i,j,k;
void ff()
{
    for(k=1; k<=n; k++)
        for(i=1; i<=n; i++)
            for(j=1; j<=n; j++)
                g[i][j]=g[i][j] || (g[i][k]&&g[k][j]);  
}
int main()
{
    int u, v;
    memset(g, 0, sizeof(g));
    scanf("%d%d", &n,&m);
    while(m--)
    {
        scanf("%d%d", &u,&v);
        g[u][v] = 1;
    }
    ff();
    int s, e=0;
    for(i=1; i<=n; i++)
    {
        s=0;
        for(j=1; j<=n; j++)
        {
            if(g[i][j] || g[j][i])
                s++;
        }
        if(s==n-1)
           e++;
    }
    printf("%d", e);
    return 0;
}

题意   一群牛比赛，每场两只牛对打，并分出胜负，现在问你能确定几只牛的战斗力排名。

思  利用Folyed 来找找出每头牛可以和多少头牛建立关系，当一头牛可以和剩下的N-1头牛都可以建立关系时，它的名次就可以                                                  确定了