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
题目思路：只要判断YES就好了。如果是YES的话，那么所有的人物所在地和所有的目的地这两个数组的内容是一样的，只是位置不一样，所以只要把他们排下序。然后两个数组的相同位置相减，当不等于0的时候结束循换输出NO。否则YES
代码如下：（本人小白，如有说出，望各位海涵）
#include <iostream>
#include <algorithm>
using namespace std;
int a[500000],b[500000];
int main()
{
int n;
while(cin>>n&&n)
{
for(int i=0; i<n; i++)
cin>>a[i]>>b[i];
int f=1;
sort(a,a+n);
sort(b,b+n);
for(int i=0; i<n; i++)
{
if(a[i]!=b[i])
{
f=0;
break;
}
}
//cout<<f<<endl;
if(f)
cout<<"YES"<<endl;
else
cout<<"NO"<<endl;
}
return 0;
}