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

題意分析：給出一系列交換位置的需求，如果同時存在(a b)交換和(b a)交換，則兩者交換成功，因此直接判斷是否存在一對交換即可。一開始使用map存儲：

map<int, int> mp;

第一次碰到a和b交換時，記錄mp[a] = b；後面碰到(b, a)交換時，匹配成功，並將其從map中刪除。但後面發現關鍵字並不唯一，比如：

1 2
1 3
3 1
2 1
這種情況下，前面的記錄會被覆蓋，因此采用multimap來存儲。

multimap<int, int> mp;

根據關鍵字從mp中查找後還要繼續遍歷找到對應的一對交換，找到則刪除，否則加入mp。

最後如果mp中為空，則表明全部交換成功，否則說明有不成功的情況。

AC源碼

 1 #include <iostream>
 2 #include <string>
 3 #include <list>
 4 #include <sstream>
 5 #include <map>
 6 using namespace std;
 7 
 8 int main()
 9 {
10 //    freopen("d:\\data1.in","r",stdin);
11     int n;
12     while(cin>>n, n)
13     {
14         multimap<int, int> mp;
15         for(int i=0;i<n;i++)
16         {
17             int x, y;
18             cin>>x>>y;
19             int num = mp.count(y);//找到元
20             multimap<int,int>::iterator it = mp.find(y);
21             int flag = 0;
22             for (int j = 0;j!=num; j++, it++)
23             {
24                 int  v = it->second;
25                 if(v==x)
26                 {
27                     mp.erase(it);
28                     flag = 1;
29                     break;
30                 }    
31             }
32             if(flag==0)
33             {
34                 mp.insert(make_pair(x, y));
35             }
36         }
37         if(mp.size()==0)
38             cout<<"YES"<<endl;
39         else
40             cout<<"NO"<<endl;
41     }
42     return 0;
43 }

紫書第五章訓練 uva 10763 Foreign Exchange by crq