Foreign Exchange UVA - 10763

  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
3
1 2
2 3
3 1
0

Sample Output

YES
NO
YES

HINT

  一个错误的题目，按照题目的意思就是只有A->B，然后B->A才是符合题意得，但实际上题目是按照只要离开学校和来到学校得一样多，最终一个学校得人数没有减少或者增加就好。

  这样直接使用map映射就可以，键就是学校，值就是人数变化。来开学校则键值--，进入学校就++，如果键值为0就删除。因此如果程序最后得map是空的说明是符合要求的。

Accepted

#include<iostream>
#include<algorithm>
#include<map>
using namespace std;
map<int, int>arr;

int main()
{
	int a, b,n;
	while (cin >> n && n)
	{
		arr.clear();
		while (n--)
		{
			cin >> a >> b;
			arr[a]--;
			if (arr[a] == 0)arr.erase(a);
			if (arr.count(b))arr[b]++;
			else arr[b] = 1;
			if (arr[b] == 0)arr.erase(b);
		}
		if (arr.empty())cout << "YES" << endl;
		else cout << "NO" << endl;
	}
}