hdu Tian Ji -- The Horse Racing

Tian Ji -- The Horse Racing

Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 65536/32768K (Java/Other)

Total Submission(s) : 66   Accepted Submission(s) : 21

Font: Times New Roman | Verdana | Georgia

Font Size: ← →

Problem Description

Here is a famous story in Chinese history.

"That was about 2300 years ago. General Tian Ji was a high official in the country Qi. He likes to play horse racing with the king and others."

"Both of Tian and the king have three horses in different classes, namely, regular, plus, and super. The rule is to have three rounds in a match; each of the horses must be used in one round. The winner of a single round takes two hundred silver dollars from the loser."

"Being the most powerful man in the country, the king has so nice horses that in each class his horse is better than Tian's. As a result, each time the king takes six hundred silver dollars from Tian."

"Tian Ji was not happy about that, until he met Sun Bin, one of the most famous generals in Chinese history. Using a little trick due to Sun, Tian Ji brought home two hundred silver dollars and such a grace in the next match."

"It was a rather simple trick. Using his regular class horse race against the super class from the king, they will certainly lose that round. But then his plus beat the king's regular, and his super beat the king's plus. What a simple trick. And how do you think of Tian Ji, the high ranked official in China?"



Were Tian Ji lives in nowadays, he will certainly laugh at himself. Even more, were he sitting in the ACM contest right now, he may discover that the horse racing problem can be simply viewed as finding the maximum matching in a bipartite graph. Draw Tian's horses on one side, and the king's horses on the other. Whenever one of Tian's horses can beat one from the king, we draw an edge between them, meaning we wish to establish this pair. Then, the problem of winning as many rounds as possible is just to find the maximum matching in this graph. If there are ties, the problem becomes more complicated, he needs to assign weights 0, 1, or -1 to all the possible edges, and find a maximum weighted perfect matching...

However, the horse racing problem is a very special case of bipartite matching. The graph is decided by the speed of the horses --- a vertex of higher speed always beat a vertex of lower speed. In this case, the weighted bipartite matching algorithm is a too advanced tool to deal with the problem.

In this problem, you are asked to write a program to solve this special case of matching problem.

Input

The input consists of up to 50 test cases. Each case starts with a positive integer n (n <= 1000) on the first line, which is the number of horses on each side. The next n integers on the second line are the speeds of Tian’s horses. Then the next n integers on the third line are the speeds of the king’s horses. The input ends with a line that has a single 0 after the last test case.

Output

For each input case, output a line containing a single number, which is the maximum money Tian Ji will get, in silver dollars.

Sample Input

392 83 7195 87 74220 2020 20220 1922 180

Sample Output

20000

题目大意：田忌和国王赛马，每个人各有n匹马，在每轮比赛中赢着得200元输者赔200元，求田忌在比赛中共获得多少钱。

解题思路：分三种情况讨论：

1：田忌最好的马比国王最好的马快，则比之。

2：田忌最好的马比国王的最好的慢，则用田忌最慢的与国王最快的比。

3：田忌最好的马和国王的马一样。可分以下三种情况讨论：

（1）：田忌最慢的马比国王最慢的马快，用慢马比国的慢马。

（2）：田忌最慢的马比国王最慢的慢，用慢的比国王最快的。

（3）：田忌最慢的马和国王最慢的马一样，且比国王最快的马慢，则用慢的比国王最快的。

代码：

#include <iostream>#include <algorithm>using namespace std;bool cmp(int a,int b){    return a>b;}int T[1010],K[1010];int main(){    int n,m,ti,tj,ki,kj,win,i,j;    while(cin>>n&&n!=0)    {        for(i=0;i<n;i++)            cin>>T[i];        for(i=0;i<n;i++)            cin>>K[i];        sort(T,T+n,cmp);        sort(K,K+n,cmp);        ti=ki=0;        tj=kj=n-1;        win=0;        while(n--)        {        if(T[ti]>K[ki])//如果田忌最快的马比齐王最快的马快        {            ti++;            ki++;            win++;        }        else            if(T[ti]<K[ki])//如果田忌最快的马比齐王最快的马慢，则用田最慢的马跟齐最快的马比            {                tj--;                ki++;                win--;            }            else//如果田忌最快的马的速度与齐威王最快的马速度相等            {                if(T[tj]>K[kj])//如果田忌最慢的比齐威王最慢的快
                {                    tj--;                    kj--;                    win++;                }                else                    if(T[tj]<K[kj])//如果田忌最慢的比齐威王最慢的慢                    {                        tj--;                        ki++;                        win--;                    }                    else                        if(T[tj]<K[ki])//田忌最慢的与齐威王最慢的相等                        {                            tj--;                            ki++;                            win--;                        }            }        }            cout<<win*200<<endl;        }        return 0;    }