poj2287 Tian Ji -- The Horse Racing（2016xynu暑期集训检测 -----C题）

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

贪心

主要是贪心策略

起初的贪心策略是 

1.对田忌和齐王的马 全部按照从快到慢的顺序排列

2.如果齐王跑的最快的马比田忌跑的最快的马慢 田忌胜

3.如果齐王跑的最快的马比田忌跑的最快的马快或相等 那么用田忌最小的马来消耗 田忌输

后来发现 

10 10 6 1

10 10 8 1

这个数据竟然为0

明明应该是200

想了好久   这种策略不对

对于3  应该修改为：

3.如果如果齐王跑的最快的马比田忌跑的最快的马快或相等 那么用田忌最小的马和齐王跑的最慢的马比较 如果田忌的小于或等于齐王的  那么直接和齐王

最快的马匹配，否则 田忌最慢的马和齐王跑的最慢的马匹配   田忌胜。

主要代码如下:

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       #include <stdio.h>
       
       

       
       #include <algorithm>
       
       

       
       using 
       
       namespace 
       
       std;
       
       

       
       bool 
       
       cmp(
       
       int 
       
       x,
       
       int 
       
       y)
       
       
{
       
       
  
       
       return 
       
       x
       
       >
       
       y;
       
       
}
       
       

       
       int 
       
       main()
       
       
{
       
       
  
       
       int 
       
       n;
       
       
  
       
       int 
       
       a[
       
       1005];
       
       
  
       
       int 
       
       b[
       
       1005];
       
       
  
       
       while(
       
       ~scanf(
       
       "%d",
       
       &
       
       n)
       
       &&
       
       n)
       
       
  {
       
       
    
       
       for(
       
       int 
       
       i
       
       =
       
       0;
       
       i
       
       <
       
       n;
       
       i
       
       ++)
       
       
    
       
       scanf(
       
       "%d",
       
       &
       
       a[
       
       i]);
       
       
    
       
       for(
       
       int 
       
       i
       
       =
       
       0;
       
       i
       
       <
       
       n;
       
       i
       
       ++)
       
       
    
       
       scanf(
       
       "%d",
       
       &
       
       b[
       
       i]);
       
       
    
       
       sort(
       
       a,
       
       a
       
       +
       
       n,
       
       cmp);
       
       
    
       
       sort(
       
       b,
       
       b
       
       +
       
       n,
       
       cmp);
       
       
    
       
       int 
       
       win
       
       =
       
       0;
       
       
    
       
       int 
       
       tie
       
       =
       
       0;
       
       
    
       
       int 
       
       lose
       
       =
       
       0;
       
       
    
       
       int 
       
       l1
       
       =
       
       0,
       
       r1
       
       =
       
       n
       
       -
       
       1;
       
       
    
       
       int 
       
       l2
       
       =
       
       0,
       
       r2
       
       =
       
       n
       
       -
       
       1;
       
       
    
       
       int 
       
       cnt
       
       =
       
       0;
       
       
    
       
       while(
       
       cnt
       
       ++<
       
       n)
       
       
    {
       
       
      
       
       //如果齐王当前最快的马小于田忌当前最快的马 田忌胜 
       
       
      
       
       if(
       
       b[
       
       l2]
       
       <
       
       a[
       
       l1])
       
       
      {
       
       
        
       
       win
       
       ++;
       
       
        
       
       l1
       
       ++;
       
       
        
       
       l2
       
       ++;
       
       
      }  
       
       
      
       
       else
       
       
      {
       
       
        
       
       //如果齐王当前最慢的马 大于等于 田忌当前最慢的马 消耗当前齐王最快的马 田忌输 
       
       
        
       
       //否则 齐王最慢的马和田忌最慢马匹配 田忌胜 
       
       
        
       
       if(
       
       b[
       
       r2]
       
       >=
       
       a[
       
       r1])
       
       
        {
       
       
          
       
       if(
       
       b[
       
       l2]
       
       ==
       
       a[
       
       r1])
       
       
          {
       
       
            
       
       tie
       
       ++;
       
       
          }
       
       
          
       
       if(
       
       b[
       
       l2]
       
       >
       
       a[
       
       r1])
       
       
          {
       
       
            
       
       lose
       
       ++;
       
       
          }
       
       
          
       
       l2
       
       ++;
       
       
        }
       
       
        
       
       else
       
       
        {
       
       
          
       
       r2
       
       --;
       
       
          
       
       win
       
       ++;
       
       
        }
       
       
        
       
       r1
       
       --;
       
       
      }  
       
       
    }
       
       
    
       
       printf(
       
       "%d\n",(
       
       win
       
       -
       
       lose)
       
       *
       
       200);
       
       
  }
       
       
  
       
       return 
       
       0;
       
       
}
      
      
      
      

       
       
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