poj3660(floyd)--Java_java poj 3660-优快云博客

本文介绍了一种通过一系列比赛结果来精确确定参赛牛只技能排名的算法。该算法使用Floyd算法遍历所有可能的关系，最终确定哪些牛的排名可以被确切地定位。

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Cow Contest

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 14363		Accepted: 8048

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

Sample Output

Source

USACO 2008 January Silver

import java.util.*;
public class Main {

    static int map[][];
    static int n,m;
    public static void floyd()
    {
        int i,j,k;
        for(k=1;k<=n;k++)
            for(i=1;i<=n;i++)
                for(j=1;j<=n;j++)
                {
                    if (map[i][j] == 0 && map[i][k]==1 && map[k][j]==1)//遍历所有传递关系
                        map[i][j]=1;
                }
    }
    public static void main(String[] args) {
        // write your code here
        Scanner in=new Scanner(System.in);
        n=in.nextInt();
        m=in.nextInt();

        map=new int[n+1][n+1];

        for(int i=0;i<m;i++)
        {
            int a=in.nextInt();
            int b=in.nextInt();

            map[a][b]=1;//如果a>b 则标记为1
        }

        floyd();
        int count=0;
        for(int i=1;i<=n;i++)
        {
            int sum=0;
            for(int j=1;j<=n;j++)
            {
                if(map[i][j]==1 || map[j][i]==1)//如果一头牛与其他n-1头牛都存在关系，则可确定排名
                {
                    sum++;
                }
            }
            if(sum==n-1)
            {
                count++;
            }
        }

        System.out.println(count);

    }

}