kuangbin 最短路 H题（如何将看似拓扑的问题转化成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 cowA has a greater skill level than cow B (1 ≤ A ≤ N; 1 ≤B ≤ N; A ≠ B), then cow A will always beat cowB.

Farmer John is trying to rank the cows by skill level. Given a list the results ofM (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 54 34 23 21 22 5

Sample Output 2

题意：告诉你N只牛之间部分的两两胜负关系，求有几只牛的位置是确定的。

思路：此题看似为拓扑，但若使用拓扑排序，如何判断位置是否确定就是件很麻烦的事情。
这里，我们可以看出，如果A赢了B，B赢了C，则A就赢了C。此处很有floyd算法中dist[i][j]=min(dist[i][j],dist[i][k]+dist[k][j])的感觉。
使用floyd算法，当一只牛与其他牛的胜负关系确定了，那么这头牛的位置也就确定了。

代码：

//By Sean Chen#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <queue>using namespace std;int Map[105][105];    //牛与牛之间的胜负关系，0表示无关系，1表示A赢B，-1表示A输Bint n,m,a,b;int main(){    scanf("%d%d",&n,&m);    for (int i=0;i<m;i++)    {        scanf("%d%d",&a,&b);        Map[a][b]=1;        Map[b][a]=-1;    }    for (int k=1;k<=n;k++)        for (int i=1;i<=n;i++)            for (int j=1;j<=n;j++)            {                if (Map[i][k]==1 && Map[k][j]==1)                    Map[i][j]=1;          //i赢j                if (Map[i][k]==-1 && Map[k][j]==-1)                    Map[i][j]=-1;         //i负于j            }    int ans=0;    for (int i=1;i<=n;i++)    {        int flag=1;        for (int j=1;j<=n && flag;j++)        {            if (i!=j && Map[i][j]==0)                flag=0;        }        ans+=flag;    }    cout<<ans<<endl;    return 0;}