UVa 10763 Foreign Exchange（水题）

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                               Output for 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

YES

NO

一个学生想去做交换生，假设有任意一个申请A到B的，但是没有B到A的申请， 那么这批申请表都不能被处理

本来是记录下各个点的入度于出度，其实没必要想那么多，定义一个数组初始为0，A到B，A这个地方就-1，代表缺一个人，B的则+1，最后看每个地方是否全为0，一遍水过~~

#include <iostream>#include <cstdio>#include <cstring>#include <stdlib.h>#define N 500010using namespace std;int a[N];int main(){    int n;    while(scanf("%d",&n),n)    {        memset(a,0,sizeof(a));        int max1=-1;        int x,y;        for(int i=0;i<n;i++)        {            scanf("%d%d",&x,&y);            a[x]--;            a[y]++;            max1=max(max1,x);            max1=max(max1,y);        }        int flag=1;        for(int i=0;i<=max1;i++)        {            if(a[i]!=0)            {                flag=0;                break;            }        }        if(flag) printf("YES\n");        else printf("NO\n");    }    return 0;}