2015 HUAS Summer Training#1~D

10763 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

解题思路：首先看到这个问题应该是是个排序问题。它也是以0为结束标志，所以每一次输入时要先判断其输入的数字是否为0。题目的意思是交换学生深造学习，第一个数字表示这个学生的目前所在地方，第二个数字表示这个学生的目标地方。其实简单来说就是把第一列和第二列分别排好序，看两列的排序结果是不是相同的，如果是，就输出"YES"，否则输出"NO"。

程序代码：

#include<cstdio> #include<algorithm> using namespace std; const int maxn=500000; int a[maxn],b[maxn]; int main() { int n,count; while(scanf("%d",&n)==1&&n) { count=0; for(int i=0;i<n;i++) scanf("%d%d",&a[i],&b[i]); sort(a,a+n); sort(b,b+n); for(int j=0;j<n;j++) if(a[j]==b[j]) count++; if(count==n) printf("YES\n"); else printf("NO\n"); } return 0; }