UVA - 10763 Foreign Exchange

Problem E
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

这道题刚开始几遍都是TLE，最后AC也有1592MS，感觉效率不高。。。

题目本身方面，要注意可能会出现两个A相同的情况，所以不能用map，而用multimap；

其他就是练习C++语法方面的了。练习STL的find、pair、insert、erase等等，熟练的话应该好做；

没有找到更好的更快的好方法。。。

#include 
using namespace std;

int main()
{
    int n;
    while(cin >> n, n!=0)
    {
        multimap table;
        multimap::iterator it;
        for(int i = 0; i < n; i++)
        {
            int A, B;
            cin >> A >> B;
            for(it = table.find(B); it!=table.end() && (*it).first==B; it++)
            {
                if((*it).second == A)
                    break;
            }
            if((*it).second == A)
                table.erase(it);
            else
                table.insert({A, B});
        }

        if(!table.size())
            printf("YES\n");
        else
            printf("NO\n");
    }
    return 0;
}

转载于:https://www.cnblogs.com/kunsoft/p/5312794.html