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): 50442    Accepted Submission(s): 21153

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

 

#include<iostream>
#include<stdio.h>
using namespace std;
const int maxx = 1005;
int main(){
    int t;
    scanf("%d",&t);
    while(t--){
        int n,v;
        scanf("%d%d",&n,&v);
        int val[maxx];
        int wei[maxx];
        for(int i=1;i<=n;i++){

            scanf("%d",&val[i]);
        }
        for(int i=1;i<=n;i++){
            scanf("%d",&wei[i]);
        }
        int dp[maxx][maxx];
        for(int i=0;i<=v;i++)
          dp[0][i]=0;
        for(int i=1;i<=n;i++){
            for(int j=0;j<=v;j++){
              
                if(j<wei[i]){
                    dp[i][j]=dp[i-1][j];
                }else{
                    dp[i][j]=max(dp[i-1][j],dp[i-1][j-wei[i]]+val[i]);
                }
            }
        }
        printf("%d
",dp[n][v]);
    }
}