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：有n头牛，m个击败关系。问你最后有多少头牛的名次是可以确定的。
解题思路：Floyd传递闭包后，判断牛i前面有多少头牛，他后边有多少头牛。如果前后牛的头数等于n-1，那么说明他是可以确定名次的。
#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<iostream>
using namespace std;
int d[300][300];
int main(){
int n,m;
while(scanf("%d%d",&n,&m)!=EOF){
int a,b;
for(int i = 0; i < m;i++){
scanf("%d%d",&a,&b);
d[a][b] = 1;
}
for(int k = 1; k <= n; k++){
for(int i = 1; i <= n; i++){
for(int j = 1; j <= n; j++){
d[i][j] = (d[i][j]|| (d[i][k]&&d[k][j]));
}
}
}
int res = 0;
for(int i = 1; i <= n; i++){
int num = 0;
for(int j = 1; j <=n; j++){
if(j == i) continue;
if(d[i][j] || d[j][i]) num++;
}
if(num == n-1) res++;
}
printf("%d
",res);
}
return 0;
}