Foreign Exchange（交换生换位置）

　　　　　　　　　　 Foreign Exchange

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

题解：开两个数组s【i】，t【i】，然后把数组数据从小到大排序，循环如果能以一一对应相等，则说明可以交换，输出YES，否则只要有一个不相等就输出NO.

#include<iostream>
#include<algorithm>
using namespace std;
int s[],t[];
//注意数组大小
int main()
{
　　int n,i;
　　while(cin>>n&&n)
{
　　for(i=; i<n; i++)
　　　　cin>>s[i]>>t[i];
　　sort(s,s+n);//排序
　　sort(t,t+n);
　　int f=;
　　for(i=; i<n; i++)
{
　　　　if(s[i]!=t[i])
　　{
　　　　　　f=;
　　　　　　break;
　　}
}

if(f)
　　　　cout<<"NO"<<endl;
else
　　　　cout<<"YES"<<endl;
}
return ;

}