Foreign Exchange UVA - 10763

제목:

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
십
1 2
2 1
3 4
4 3
100 200
200 100
57 2
2 57
1 2
2 1
십
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

제목:

너에게 숫자 n을 하나 주면 n개의 데이터가 있다는 것을 대표한다. 모든 데이터는 두 개의 숫자 a와 b이다. 학생 a가 학생 b와 교환해야 한다는 것을 대표한다. 학생 b도 학생 a와 교환해야 할 때 교환에 성공할 수 있다.모든 학생이 교환에 성공하면 YES를 출력하고 그렇지 않으면 NO를 출력한다.
n==0 시 데이터의 입력이 끝나는 것을 나타낸다.

생각:

한 개의 수조로 학생들이 교체를 요구하는 정보를 기록한 다음에 판단한다.
a [ x ][ y ] = s;//학생 x를 대표하여 학생 y와 교환하고 싶은 것은 s개이다.

코드는 다음과 같습니다.

#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])// ；
            return 0;
    }
    return 1;
}

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