POJ3660 Cow Contest【最短路-floyd】

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

思路：原来是觉得像昨天写的那题一样用bellman判断一下能不能成环

但是这道题问的不只是一只牛 而是所有牛 而且成环不成环并不能判断他的排名能不能确定

对于一头牛 如果我们知道有x只牛能赢他 他能赢y只牛 并且x+y=n-1的话 那么他的排名是可以确定的

所以就用floyd跑一遍 就可以把传递的关系建立起来了

emmmm传递闭包？题解上有这么说 但是不是很理解有什么关系......因为用了floyd？？？

唉离散学了都忘光了 感觉之前学的真的都忘了

然后遍历每一头牛看看行不行

代码：

#include<stdio.h>
#include<iostream>
#include<algorithm>
#include<cmath>
#include<map>
#include<cstring>
#include<queue>
#include<stack>
#define inf 0x3f3f3f3f
using namespace std;
int n, m, d[105][105];
void floyd()
{
for(int k = 1; k <= n; k++){
for(int i = 1; i <= n; i++){
for(int j = 1; j <= n; j++){
if(d[i][k] && d[k][j]){
d[i][j] = 1;
}
}
}
}
}
int main()
{
while(cin>>n>>m){
for(int i = 0; i < m; i++){
int a, b;
cin>>a>>b;
d[a][b] = 1;
}
floyd();
int ans = 0;
for(int i = 1; i <= n; i++){
int num = 0;
for(int j = 1; j <= n; j++){
if(d[i][j] || d[j][i]){
num++;
}
}
if(num == n - 1){
ans++;
}
}
cout<<ans<<endl;
}
return 0;
}