D - 交换学生

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

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

Sample Output

YES

NO

#include<iostream>
#include<algorithm>
#include<vector>
#include<cstdio>
#include<stdio.h>
#include<string>
#include<cstring>
using namespace std;
int main()
{
	int n;
	while(cin >> n && n!=0)
	{
		int c[500006];
		memset(c,0,sizeof(c));
		int a,b,i,k;
		for(i=0;i<n;i++)
		{
			cin >> a >> b;
			c[a]+=1;c[b]+=1;
		}
		k=1;
		for(i=0;i<500005;i++)
		{
			if(c[i]%2!=0)//若是有一对的人，那么对应的数一定是偶数
			{k=0;break;}
		}
		if(k==1)
		printf("YES\n");
		else
		printf("NO\n");
	}
	return 0;
}

但是本人这么做跟题意有不符，题目要求如果有人从A到B，一定要有人从B到A才能交换，

举例子：1，2      2，3     3，1     

按照题目，这三人是不能交换的，输出NO，而我的是YES，不过我交到vjudge上仍然是对的（滑稽笑）