Uva 10763 Foreign Exchange

Foreign Exchange
Input: standard input
Output: standard output
Time Limit: 1 second

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

题意：给出N个数量的每个人的出发地和目的地，如果每个人的出发地都能找到另一个人作为目的地，目的地是另一个人的出发地，就算配对成功，所有人都配对成功就输出YES，否则输出NO。

分析：其实将目的地和出发地都排一个序，然后比较是否都相等，相等则输出YES，不相等则输出NO。

不过后来我发现，这样的做法有一个漏洞，就是如果数据是 1->2 ,2->3, 3->4, 4->1  ,  1->3, 2->4, 3->1 , 4->2，这个并不能满足题目所说的一个A->B必有一个B->A的意思。数据水了，但是我还是改成了用矩阵邻接表的写法做这个题（还是按照题目要求来）。不过数据还是挺水的，数据量不大。

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#define MAX_N 5005using namespace std;int v[MAX_N][MAX_N];int main(){    int n;    while(~scanf("%d",&n)&&n)    {        memset(v,0,sizeof(v));        int x,y;        for(int i=1;i<=n;i++)        {            scanf("%d%d",&x,&y);            v[x][y]++;            v[y][x]--;        }        int flag=1;        for(int i=1;i<=1000;i++)        {            for(int j=1;j<=1000;j++)            {                if(v[i][j]!=0)                {                    flag=0;                    break;                }            }        }        if(flag) printf("YES\n");        else   printf("NO\n");    }    return 0;}