poj-3660-cows contest（不懂待定）

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
分析：有人说叫闭包传递。这题就是用一种类似于floyd的算法，
开始时，如果a胜b则由a到b连一条边。这样所有a能走到的点都是排名在a以后的。
所有能走到a的点都是排名在a以前的。用floyd，求出每个点能到达的点。
如果有一个点，排名在它之前的和排名在它之后的点之和为n-1，那么它的排名就是确定的。

#include<iostream>

using namespace std ;

int main()

{

    int N,M,a,b;

    cin>>N>>M;

    int aa[110][110]={0};

      while(M--)

    {

        cin>>a>>b;

        aa[a][b]=1;

    }

    for(int i=1;i<=N;i++)

        for(int j=1;j<=N;j++)

        for(int k=1;k<=N;k++)

    {

        if(aa[j][i]&&aa[i][k])

            aa[j][k]=1;

    }

    int ans=0;

    for(int i=1;i<=N;i++)

    {

        int tmp=0;

        for(int j=1;j<=N;j++)

        tmp+=aa[i][j]+aa[j][i];

        if(tmp==N-1)ans++;

    }

  cout<<ans<<endl;

    return 0;

}

巴特西

poj-3660-cows contest（不懂待定）

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