Bone Collector 01背包

                                         Bone Collector

                    Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
                    Total Submission(s): 21711 Accepted Submission(s): 8752

Problem Description

Many years ago , in Teddy’s hometown there was a man who was called “Bone Collector”. This man like to collect varies of bones , such as dog’s , cow’s , also he went to the grave …
The bone collector had a big bag with a volume of V ,and along his trip of collecting there are a lot of bones , obviously , different bone has different value and different volume, now given the each bone’s value along his trip , can you calculate out the maximum of the total value the bone collector can get ?

Input

The first line contain a integer T , the number of cases.
Followed by T cases , each case three lines , the first line contain two integer N , V, (N <= 1000 , V <= 1000 )representing the number of bones and the volume of his bag. And the second line contain N integers representing the value of each bone. The third line contain N integers representing the volume of each bone.

Output

One integer per line representing the maximum of the total value (this number will be less than 2 ³¹).

Sample Input

   
   
   
   

    
    
    
    1
5 10
1 2 3 4 5
5 4 3 2 1
   
   
   
   

Sample Output

   
   
   
   

    
    
    
    14
   
   
   
   

Author

Teddy

Source

HDU 1st “Vegetable-Birds Cup” Programming Open Contest

Recommend

lcy

code:

#include<iostream>
#include<string.h>
#include<cstdio>
using namespace std;
int main()
{
    int cas;
    int value[1001],weight[1001];
    int num,totalweight;
    int f[1001];
    cin>>cas;
    while(cas--)
    {
        cin>>num>>totalweight;
        memset(weight,0,sizeof(weight));
        memset(value,0,sizeof(value));
        memset(f,0,sizeof(f));
        for(int i=0;i<num;i++)
        {
            cin>>value[i];
        }
        for(int i=0;i<num;i++)
        {
            cin>>weight[i];
        }
        
        f[0]=0;
        for(int i=0;i<num;i++)
         for(int v=totalweight;v>=weight[i];v--)
         {
             f[v]=max(f[v],f[v-weight[i]]+value[i]);
         }
         cout<<f[totalweight]<<endl;
     
        
    }
        return 0;
}