poj 3660 Cow Contest ([kuangbin带你飞]专题四 最短路练习)

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

题目大意：给出牛之间的强弱关系，让你确定有多少头牛能够确定其排名.

解题思路：只要知道眉头牛和其他牛的强弱关系就可以知道这个牛的强弱，，

#include
#include
#include
using namespace std;
int n,m;
const int maxn = 100+10;
int mp[maxn][maxn];

int main(){
    while(scanf("%d%d",&n,&m)!=EOF){
        memset(mp,0,sizeof(mp));
        for(int i=1;i<=m;i++){
            int a,b;
            scanf("%d%d",&a,&b);
            mp[a][b]=1;
        }
    for(int k=1;k<=n;k++){
        for(int i=1;i<=n;i++){
            for(int j=1;j<=n;j++){
                    if(mp[i][k] && mp[k][j]) mp[i][j]=1;
                }
            }
        }
        int ans=0;
        for(int i=1;i<=n;i++){
            int res= n-1;
            for(int j=1;j<=n;j++){
                if(mp[i][j] || mp[j][i]) res--;
            }
            if(!res) ans++;
        }
        printf("%d\n",ans);
    }





    return 0;
}