Foreign Exchange UVA - 10763

题目：

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：

给你一个数字n，代表有n个数据，每一个数据都是两个数字 a和b，代表学生a要和学生b交换，当学生b也要和学生a进行交换的时候才可以成功进行交换；当所有的学生都成功的进行了交换就输出YES，否则输出NO；

当n==0时代表数据的输入结束；

思路：

用一个数组来记录学生要求更换的信息，然后进行判断；

a [ x ][ y ] = s;//代表学生x想要和学生y进行交换的有s个；

代码如下：

#include
#include
#include
#include
using namespace std;

const int N=1010;

struct node
{
    int x;
    int y;
};

int a[N][N];
int n;

queueq;

void init()
{
    memset(a,0,sizeof a);
    while(!q.empty())
        q.pop();
    return ;
}

int panduan()
{
    node p;
    while(!q.empty())
    {
        p=q.front();
        q.pop();
        if(a[p.x][p.y]!=a[p.y][p.x])//交换不对称；
            return 0;
    }
    return 1;
}

int main()
{
    while(~scanf("%d",&n)&&n)
    {
        init();
        node p;
        for(int i=0;i