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

Sample Output

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;

}

巴特西

POJ3660 Cow Contest【最短路-floyd】

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