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骑猪看日落 2022-06-12 05:49 262阅读 0赞

Your non-profit organization (iCORE - international Confederation of Revolver Enthusiasts) coordinatesa very successful foreign student exchange program. Over the last few years, demand hassky-rocketed and now you need assistance with your task.The program your organization runs works as follows: All candidates are asked for their originallocation and the location they would like to go to. The program works out only if every student has asuitable exchange partner. In other words, if a student wants to go from A to B, there must be anotherstudent 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!InputThe input file contains multiple cases. Each test case will consist of a line containing n – the numberof candidates (1 ≤ n ≤ 500000), followed by n lines representing the exchange information for eachcandidate. Each of these lines will contain 2 integers, separated by a single space, representing the candidate’soriginal location and the candidate’s target location respectively. Locations will be representedby nonnegative integer numbers. You may assume that no candidate will have his or her original locationbeing 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.OutputFor 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

1 2

2 1

3 4

4 3

100 200

200 100

57 2

2 57

1 2

2 1

1 2

3 4

5 6

7 8

9 10

11 12

13 14

15 16

17 18

19 20

Sample Output

YES

/*
    有两个人A和B,他们各有n个数，如果A拥有的数和B拥有的数全都一样的话就输出YES
    */

#include <cstdio>
    #include <cstring>
    #include <algorithm>
    using namespace std;
    int a[500001],b[500001];
    int main()
    {
        int n;
        while(~scanf("%d",&n),n)
        {
            int i;
            for(i=0;i<n;i++)
                scanf("%d %d",&a[i],&b[i]);
            sort(a,a+n);
            sort(b,b+n);
            for(i=0;i<n;i++)
            {
                if(a[i]!=b[i])
                    break;
            }
            if(i==n)
                printf("YES\n");
            else
                printf("NO\n");
        }
        return 0;
    }
    /*
    有两个人A和B,他们各有n个数，如果A拥有的数和B拥有的数全都一样的话就输出YES
    */