HDOJ 1052 Tian Ji -- The Horse Racing

Tian Ji -- The Horse Racing

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 23508    Accepted Submission(s): 6876

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
   
   
   
   

 

田忌赛马

 

大致意思：

田忌与齐王赛马，每回合输者给赢者200银元，试编出程序，使田忌赢的钱尽可能多。

输入：

每组数的第一行为比赛的马的个数（<=100）第二行为田忌的马的速度，第三行为齐王的马的速度。

输出：

田忌赢得的钱数

 

思路：

1、田忌的当前最快的马（速度）>齐王当前最快的马：

   直接比，能赢干嘛不赢呀。

2、田忌的当前最快的马<齐王当前最快的马：

   田忌最慢的马跟齐王最快的马。

3、田忌的当前最快的马=齐王当前最快的马：

      1 田忌当前最慢的马>齐王当前最慢的马：

                两个最慢的马比。

     2田忌当前最慢的马<齐王当前最慢的马：

                用田忌最慢的马跟齐王最快的马比。

     3田忌当前最慢的马=齐王当前最慢的马：

                1 、田忌当前最慢马=齐王当前最快马：

                      说明他们剩下的等数量的马都一样快，退出循环。

               2、  否则用田忌的慢马与齐王的快马比。

<程序中的if else 就是按这个顺序来的：）>

#include  
#include  
  
int a[1001]={0};//田忌的马  
int b[1001]={0};//齐王的马  
  
int comparator(const void *elem1,const void *elem2)  
{  
    return *(int *)elem2-*(int *)elem1;  
}  
  
void main()  
{  
   int n,th,ts,qh,qs,win;//th：田的快马，即序列头指针ts：田的慢马即序列尾指针，另外两个为齐的  
   while(scanf("%d",&n)&&n)  
   {  
      for(int i=0;i
          scanf("%d",&a[i]);  
      for(i=0;i
          scanf("%d",&b[i]);  
      qsort(a,n,sizeof(int),comparator);  
      qsort(b,n,sizeof(int),comparator);//快排  
      th=qh=win=0;  
      ts=qs=n-1;  
      while(th<=ts)//当田所有的马遍历一遍时，退出循环。  
      {  
          if(a[th]>b[qh])  
          {  
              th++;  
              qh++;  
              win++;  
          }  
          else if(a[th]
          {  
              ts--;  
              qh++;  
              win--;  
          }  
          else  
          {  
              if(a[ts]>b[qs])  
              {  
                  ts--;  
                  qs--;  
                  win++;  
              }  
              else if(a[ts]
              {  
                  ts--;  
                  qh++;  
                  win--;  
              }  
              else  
              {  
                 if(a[th]==b[qs])  
                     break;  
                 else  
                 {  
                    ts--;  
                    qh++;  
                    win--;  
                 }  
              }  
          }//else  
      }//while  
      printf("%d\n",win*200);  
   }  
     
}