ACM交换生问题

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10763 Foreign Exchange

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

#include <iostream>
#include <algorithm>
using namespace std;
const int maxn=500005;
int a[maxn],b[maxn];
int main()
{
    int n;
    while(cin>>n&&n)
    {
        int k=1;
        for(int i=0;i<n;i++)
        {
            cin>>a[i]>>b[i];
        }
        sort(a,a+n);
        sort(b,b+n);
        for(int j=0;j<n;j++)
        {
            if(a[j]!=b[j])
            {
                k=0;
                break;
            }
        }
        if(k)
            cout<<"YES"<<endl;
        else
            cout<<"NO"<<endl;
    }
    return 0;
}

View Code

17 18

19 20

0

Sample Output

YES

NO

解题思路：我们先定义两个足够大的数组a和b，分别用来存放学生的起始地和目的地。输入需要的案例数，输入对应的目的地和起始地，然后对这两个数组用sort()函数（sort函数包含在头文件algorithm中）进行排序。然后将这两个数组进行比较，只有当这两个数组具有的元素全部都相等时才能进行交换（即a中与b中对应的元素相同）最后进行题目要求的输出就行了。

程序代码：

ACM交换生问题

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原文地址：http://www.cnblogs.com/xinxiangqing/p/4668310.html