杭电1052--Tian Ji -- The Horse Racing （贪心）

Tian Ji -- The Horse Racing

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 21416    Accepted Submission(s): 6278

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

3 
92 83 71 
95 87 74 
2 
20 20 
20 20 
2 
20 19 
22 18
0

 

 

Sample Output

200 
0 
0

 

 

Source

2004 Asia Regional Shanghai

 

1、如果田忌最快的马比齐王最快的马快，则赢；

2、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消耗； //这是贪心的第一步

3、如果田忌最快的马的速度与齐威王最快的马速度相等

3.1、如果田忌最慢的比齐威王最慢的快，则赢；                    //这是贪心的第二步

3.2、如果田忌最慢的比齐威王最慢的慢，田忌慢消耗齐王快；

3.3、田忌最慢的与齐威王最慢的相等，田忌慢消耗齐王快；

#include <stdio.h>
#include <algorithm>
int tian[1010], king[1010] ;
using namespace std ;

bool cmp(int a, int b)
{
    return a > b ;
}

int main()
{
    int i,n ;
    while(~scanf("%d",&n),n)
    {
        for(i=0; i<n; i++)
        scanf("%d", &tian[i]) ;
        for(i=0; i<n; i++)
        scanf("%d", &king[i]) ;
        sort(tian, tian+n, cmp) ;
        sort(king, king+n, cmp) ;    
        
        int ti, ki, tj, kj, win=0 ;
        ti=ki=0;      //记录当前速度最大马在数组中位置； 
        tj=kj=n-1 ;   //--------速度最小--------------； 
        for(i=0,win=0; i<n; i++)
        {
            if(tian[ti] > king[ki])   // 最大的比较 ；直接赢了 ； 
            {
                ti++ ;
                ki++ ;
                win++ ;    
            }    
            else if(tian[ti] < king[ki])  //最大的比较，要输——————（最小的）消耗最大的（输得要有价值）； 
            {
                tj-- ;
                ki++ ;
                win-- ;
            }
            else                     //最大的比较 ； 相等了 ； 
            {
                if(tian[tj] > king[kj])  //比较最小的 ； 直接赢了 ； 
                {
                    tj-- ;
                    kj-- ;
                    win++ ;
                }
                
                else if(tian[tj] < king[kj])  //比较最小的， 要输———消耗最大的 ； 
                {
                    tj-- ;
                    ki++ ;
                    win-- ;
                 }
                 
                 else          //最小的也相等 ； 比较 T 最小的和  K 最大的  （不存在 T 最小的比 K 最大的还大（排过虚的&&最大的两只相等））； 
                 {
                     if(tian[tj] < king[ki])   // 消耗最大的； 
                       {
                        tj-- ;
                        ki++ ;
                        win--;
                    }    
                    
                    // //相等的就不用考虑了； T最大==K最大 ； T最小==K最大； 全为平局（不管是中间开始，还是从头开始；）； 
                }
            }
        } 
        printf("%d
",win*200) ;
        
    }    
    return 0 ;
}