hdu 1052 Tian Ji -- The Horse Racing 贪心算法

Tian Ji -- The Horse Racing

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 11496    Accepted Submission(s): 3218

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

 

典型的贪心算法~~自己木有考虑到贪心的第二步导致wa了好多次

算法分析 

Problem Description：给出2N组数据，分别表示田忌和齐威王的N匹马的速度，没进行一场比赛（每组数据共N场场赛），若能分出胜负，则输的一方要给赢的一方200Y￥（银元），求田忌以怎样的策略才能赚取最多的老婆本。

Solution：这题有多种解体思路，DP，二分图最大匹配算法等，这里给出的是比较容易理解的贪心算法，具体思路如下：

先将田忌跟齐王的马的速度数组进行一次冒泡排序

1、如果田忌最快的马比齐王最快的马快，则比之

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  //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，则比之                         //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VS齐王快

3.3、田忌最慢的与齐威王最慢的相等，田忌慢VS齐王快

注意这两种 数据

3

1 2 3

928371 tian 第一步 tian1对king1 转到2 贪心的第一次，

95 9274king 第二步tian1对king2 转到3.1 （贪心的第二次 保留tian的快马 ）

                 第三步 只剩下一种选择了~~

3
92 83 70
92 91 60

#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;    //   记录 king  比赛用的马  循环跳出的判定条件 
      i=j=sum=0;
      k=n-1;
      f=n-1;
      while(1)
      {           
          if(ji==n)   break;   //   king  的马全部比完后跳出 
          if(b[j]>a[i]) {   sum-=200;j++;k--;ji++; continue;}  //如果king的比tian的快马快 用tian的慢马对king的快马 
          if(b[j]==a[i]){                                       //如果相等 
                            if(b[f]<a[k]){f--;k--;sum+=200;ji++;continue;} //看两人的慢马 tian的慢马比king的慢马快则比 
                            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的快马快 直接比 
      }          
      printf("%d\n",sum);    
   }
}