交换学生(Foreign Exchange)

Problem E
Foreign Exchange
Input: standard input
Output: standard output
Time Limit: 1 second

Your non-profit organization (iCORE - international Confederation of Revolver Enthusiasts) coordinates a very successful foreign student exchange program. Over the last few years, demand has sky-rocketed and now you need assistance with your task.

The program your organization runs works as follows: All candidates are asked for their original location and the location they would like to go to. The program works out only if every student has a suitable exchange partner. In other words, if a student wants to go from A to B, there must be another student who wants to go from B to A. This was an easy task when there were only about 50 candidates, however now there are up to 500000 candidates!

Input

The input file contains multiple cases. Each test case will consist of a line containing n - the number of candidates (1≤n≤500000), followed by n lines representing the exchange information for each candidate. Each of these lines will contain 2 integers, separated by a single space, representing the candidate's original location and the candidate's target location respectively. Locations will be represented by nonnegative integer numbers. You may assume that no candidate will have his or her original location being the same as his or her target location as this would fall into the domestic exchange program. The input is terminated by a case where n = 0; this case should not be processed.

 

Output

For each test case, print "YES" on a single line if there is a way for the exchange program to work out, otherwise print "NO".

 

Sample Input                               Output for Sample Input

10 1 2 2 1 3 4 4 3 100 200 200 100 57 2 2 57 1 2 2 1 10 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 0

YES NO

【分析】

       首先我所能想到的是（可惜该方法时间超限），先把所有的学生的A和B（出发地和目的地）都储存在二维数组中（第一列存A，第二列存B），然后在创建另个用于标记的数组，该数组标记当前行是否被访问过，假如已被访问说明该学生已经匹配到“对象”，就不能再帮他寻找“对象“了。从二维数组中的第一行开始进行匹配……

用java语言编写程序，代码如下：

import java.util.Scanner;
//Time limit exceeded
public class Main {
	public static void main(String[] args) {
		Scanner input = new Scanner(System.in);
		
		while(input.hasNext()) {
			int n = input.nextInt();
			if(n == 0)
				break;
			
			
			int[][] es = new int[n][2];
			boolean[] isVisited = new boolean[n];//默认值为false
			
			for(int i = 0; i < n; i++) {
				es[i][0] = input.nextInt();
				es[i][1] = input.nextInt();
			}
			
			if(n % 2 != 0) {
				System.out.println("NO");
				continue;
			}
			
			boolean result = true;
			for(int i = 0; i < n - 2; i++) {
				int j;
				if(isVisited[i])
					continue;
				for(j = i + 1; j < n; j++) {
					if(!isVisited[j] && es[i][0] == es[j][1] && es[i][1] == es[j][0]) {
						isVisited[j] = true;
						break;
					}	
				}
				if(j == n) {
					result = false;
					break;
				}
				//isVisited[i] = true;
			}
			
			if(result)
				System.out.println("YES");
			else
				System.out.println("NO");
		}
	}
}

       另外一种思路是这样的，在读取每个学生的A和B的过程中，将A分在一个数组中、将B分在一个数组中（数组初始化为0），这个两个数组分别以A、B为下标进行自增，每次读取到A或B下标就对相应的数组的A或B下标的元素值加1。我们只要判断这两个数组对应下标的元素值是否相等即可。

用java语言编写程序，代码如下：

import java.util.Arrays;
import java.util.Scanner;

public class Main {
	public static void main(String[] args) {
		Scanner input = new Scanner(System.in);
		int[] f = new int[500010];
		int[] t = new int[500010];
		
		while(input.hasNext()) {
			int n = input.nextInt();
			
			if(n == 0)
				break;
			
			Arrays.fill(f, 0);
			Arrays.fill(t, 0);
			
			int max = 0;
			for(int i = 0; i < n; i++) {
				int A = input.nextInt();
				int B = input.nextInt();
				if(max < A) max = A;
				if(max < B) max = B;
				f[A]++;
				t[B]++;
			}
			
			boolean result = true;
			for(int i = 0; i <= max; i++) {
				if(f[i] != t[i]) {
					result = false;
					break;
				}
			}
			if(result)
				System.out.println("YES");
			else
				System.out.println("NO");
		}
	}
}