POJ 3660 Cow Contest（传递闭包floyed算法）

Cow Contest

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 5989		Accepted: 3234

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

Source

USACO 2008 January Silver

 

 

题目给出了m对的相对关系，求有多少个排名是确定的。

使用floyed求一下传递闭包。如果这个点和其余的关系都是确定的，那么这个点的排名就是确定的。

//============================================================================

// Name        : POJ.cpp

// Author      : 

// Version     :

// Copyright   : Your copyright notice

// Description : Hello World in C++, Ansi-style

//============================================================================

/*

 * floyed算法，传递闭包。如果一个点和其余点的关系都是确定的，则这个的排名是确定的

 */

#include <iostream>

#include <string.h>

#include <algorithm>

#include <stdio.h>

using namespace std;

const int MAXN=110;

int win[MAXN][MAXN];

int main()

{

    int n,m;

    while(scanf("%d%d",&n,&m)==2)

    {

        memset(win,0,sizeof(win));

        int u,v;

        while(m--)

        {

            scanf("%d%d",&u,&v);

            win[u][v]=1;

        }

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

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

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

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

                        win[i][j]=1;

        int ans=0;

        int j;

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

        {

            for(j=1;j<=n;j++)

            {

                if(i==j)continue;

                if(win[i][j]==0&&win[j][i]==0)break;//关系不确定

            }

            if(j>n)ans++;

        }

        printf("%d\n",ans);

    }

    return 0;

}