hdu 1009 FatMouse'Trade 贪心算法 之 部分背包问题

FatMouse' Trade

Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 65536/32768K (Java/Other)

Total Submission(s) : 14   Accepted Submission(s) : 3

Problem Description

FatMouse prepared M pounds of cat food, ready to trade with the cats guarding the warehouse containing his favorite food, JavaBean.
The warehouse has N rooms. The i-th room contains J[i] pounds of JavaBeans and requires F[i] pounds of cat food. FatMouse does not have to trade for all the JavaBeans in the room, instead, he may get J[i]* a% pounds of JavaBeans if he pays F[i]* a% pounds of cat food. Here a is a real number. Now he is assigning this homework to you: tell him the maximum amount of JavaBeans he can obtain.

 

Input

The input consists of multiple test cases. Each test case begins with a line containing two non-negative integers M and N. Then N lines follow, each contains two non-negative integers J[i] and F[i] respectively. The last test case is followed by two -1's. All integers are not greater than 1000.

 

Output

For each test case, print in a single line a real number accurate up to 3 decimal places, which is the maximum amount of JavaBeans that FatMouse can obtain.

 

Sample Input

   
   
   
   

    
    
    
    5 3
7 2
4 3
5 2
20 3
25 18
24 15
15 10
-1 -1

   
   
   
   

 

Sample Output

   
   
   
   

    
    
    
    13.333
31.500

   
   
   
   
   
   
   
   

    
    
    
     
   
   
   
   
   
   
   
   

    
    
    
     
   
   
   
   
   
   
   
   

    
    
    
     
   
   
   
   
   
   
   
   

    
    
    
     
   
   
   
   
   
   
   
   

    
    
    
    首先求出每个房间里单位重量的猫粮能够换多少javabean，假设为c[i]，然后对c[i]排序，取c[i]大的数据，这样就能保证换取的javabean最多。
   
   
   
   
   
   
   
   

    
    
    
    AC代码如下：
   
   
   
   
   
   
   
   

    
    
    
    

        
    
    
    
    
    
 
     
#include<stdio.h>
int main()
{
    int i,m,n,a[1000],b[1000],k,s,j;
    double c[1000],sum,t;
    while(scanf("%d %d",&m,&n)!=EOF)
    {
        if(m==-1&&n==-1) break;    /*结束*/
        for(i=0;i<n;i++)
        {
            scanf("%d %d",&a[i],&b[i]);
            c[i]=(double)a[i]/b[i];   /*整型换成double型*/
        }
        for(i=0;i<n;i++)
         for(j=i+1;j<n;j++)
         if(c[i]>c[j])
         {
           t=c[i];
           c[i]=c[j];
           c[j]=t;
           k=b[i];
           b[i]=b[j];
           b[j]=k;
           s=a[i];
           a[i]=a[j];
           a[j]=s;
        }   /*对c[i]排序，排序的同时a[i],b[i]也要跟着变变下标*/
        sum=0.0;
        for(i=n-1;i>=0;i--)
        {
         if(b[i]<m)
            sum=sum+a[i];  /*全部换成a[i]*/
         else          
            sum+=m*c[i];  /*换一部分*/
            m=m-b[i];  /*m减小*/
           if(m<=0) break;  /*猫粮用完，跳出循环*/   
       }
    printf("%.3lf\n",sum);
 }
    return 0;
}