HDU 1.3.2 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): 4215 Accepted Submission(s): 1021

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

 

正確方法之一：

貪心法求解。

如果田忌最強的馬能賽過國王最強的馬，那麼賽一場。

如果田忌最弱的馬能賽過國王最弱的馬，那麼賽一場。

否則，用田忌最弱的馬和國王最強的馬賽一把。

鄙人的求解思路：

把國王和田忌全部的馬都排序，實力升序。

從田忌最弱的馬開始考慮，如果能賽過國王的一部分馬，那就和這一部分中最強的賽。

否則田忌這匹最弱的馬和國王最強的馬賽一把。

可是不知爲何這算法一直WA。。求讀者高見，感激不盡

下位正確做法代碼:

#include<stdio.h>
#include<iostream>
#include<algorithm>
using namespace std;
int a[3000],b[3000];
int cmp(int a,int b)
{
  return a>b;    
}
int main()
{
   int i,n,j,sum,k,f,ji;
   while( scanf("%d",&n) && n!=0 )
   {
      for(i=0;i<n;i++)
        scanf("%d",&a[i]);  
      for(i=0;i<n;i++)      
        scanf("%d",&b[i]);
      sort(a,a+n,cmp);   
      sort(b,b+n,cmp);
      ji=0;    //   記錄 king  比賽用的馬  循環跳出的判定條件 
      i=j=sum=0;
      k=n-1;
      f=n-1;
      while(1)
      {           
          if(ji==n)   break;   //   king  的馬全部比完後跳出 
          if(b[j]>a[i]) {   sum-=200;j++;k--;ji++; continue;}  //如果king的比tian的快馬快 用tian的慢馬對king的快馬 
          if(b[j]==a[i]){                                       //如果相等 
                            if(b[f]<a[k]){f--;k--;sum+=200;ji++;continue;} //看兩人的慢馬 tian的慢馬比king的慢馬快則比 
                            if(b[j]>a[k]){sum-=200;k--;j++;ji++;}
                            else {k--;j++;ji++;}
                            continue;
                        }
          if(b[j]<a[i]){sum+=200;j++;i++;ji++;continue;}  //如果tian的比king的快馬快 直接比 
      }          
      printf("%d\n",sum);    
   }
}