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2013
11-13

POJ 3660 Cow Contest [解题报告] Java

Cow Contest

问题描述 :

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 ≤ AN; 1 ≤ BN; AB), 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.

输入:

* 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

输出:

* Line 1: A single integer representing the number of cows whose ranks can be determined
 

样例输入:

5 5
4 3
4 2
3 2
1 2
2 5

样例输出:

2

解题代码:

//* @author popop0p0popo
import java.util.*;
import java.io.*;

public class Main{
 public static void main(String[] args){
  Scanner in=new Scanner(new BufferedReader(new InputStreamReader(System.in)));
		int n=in.nextInt();
		int m=in.nextInt();
		int[][] p=new int[n][n];
		for (int i=0;i< n ;i++ ){
			for (int j=0;j< n ;j++ ){
				p[i][j]=101;
			}
			p[i][i]=0;
		}
		for (int i=0;i< m ;i++ ){
			p[in.nextInt()-1][in.nextInt()-1]=1;
		}
		for (int k=0;k< n ;k++ ){
			for (int i=0;i< n ;i++ ){
				for (int j=0;j< n ;j++ ){
				 p[i][j]=getMin(p[i][j],p[i][k]+p[k][j]);
				}
			}
		}
		int w=0;
		for (int i=0;i< n ;i++ ){
			int t=0;
			for (int j=0;j< n ;j++ ){
				if (p[i][j]< 101||p[j][i]< 101){
					t++;
				}
			}
			if (t==n){
				w++;
			}
		}
		System.out.println(w);
	}

	public static int getMin(int a,int b){
		if (a>=b){
			return b;
		}
		else{
			return a;
		}
	}
}