HDU 2602 Bone Collector(01背包)

Bone Collector

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 44243    Accepted Submission(s): 18434

 

 

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 2³¹).

 

 

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

 

 

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
int dp[1010];
struct node
{
    int value;
    int volume;
}m[1010];
int main()
{
    int t;
    cin>>t;
    int n,v;
    while(t--){
        cin>>n>>v;
        memset(dp,0,sizeof(dp));
        for(int i=1;i<=n;i++) cin>>m[i].value;
        for(int i=1;i<=n;i++) cin>>m[i].volume;
        for(int i=1;i<=n;i++){
            for(int j=v;j>=m[i].volume;j--){
                dp[j]=max(dp[j],dp[j-m[i].volume]+m[i].value);
            }
        }
        cout<<dp[v]<<endl;
    }    
    return 0;
}