UVA - 10763 Foreign Exchange

Description

Your non-profit organization (iCORE - internationalConfederation 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 to500000 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

101 22 13 44 3100 200200 10057 22 571 22 1101 23 45 67 89 1011 1213 1415 1617 1819 200

Sample Output

YESNO

分析：每组数据即是一个元组，用pair存，读取一个元组就在map中映射一次，如果反向元存在则加1，否则如果正向元组存在则减1，如果两者都不存在则建立新的映射，并把ID记为1；存完了就开始循环找map中是否存在奇数或者小于1的数；

#include <iostream>#include <algorithm>#include <utility>#include <cstdio>#include <map>#include <vector>using namespace std;typedef pair<int,int> Student;Student stu;Student stub;map<Student,int> IDcache;vector<Student> Stucache;int main(){int n;while(~scanf("%d",&n)&&n){IDcache.clear(); Stucache.clear();int first,second;        for(int i=0; i<n; i++)        {scanf("%d%d",&first,&second);stu = make_pair(first,second);stub = make_pair(second,first);if(IDcache.count(stub))IDcache[stub]++;else if(IDcache.count(stu))IDcache[stu]--;else {Stucache.push_back(stu); IDcache[stu]=1;}        }int yes=1;for(int i=0; i<Stucache.size(); i++){int id = IDcache[Stucache[i]];if(id<1 || id%2){yes=0;break;}}printf("%s\n",yes?"YES":"NO");}    return 0;}