hduoj 2602Bone Collector

Bone Collector

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 40598    Accepted Submission(s): 16872

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

 

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#include<cstdio>
#include<cstring>
#include<iostream>
#include<stack>
#include<set>
#include<map>
#include<queue>
#include<algorithm>
using namespace std;
#define max(a,b) (a>b?a:b)
int va[1005],vo[1005],dp[1005][1005];
int main()
{
    //freopen("D:\INPUT.txt","r",stdin);
    int t,i,j;
    int n,v;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d %d",&n,&v);
        for(i=1; i<=n; i++)
        {
            scanf("%d",&va[i]);
        }
        for(i=1; i<=n; i++)
        {
            scanf("%d",&vo[i]);
        }
        //dp[i][j]  前i件物品放入j体积的价值的最大值
        //dp[i][j]=max(dp[i-1][j],dp[i-1][j-vo[i]]+va[i])
        for(i=1; i<=n; i++) //i体积
        {
            for(j=0; j<=v; j++)
            {
                if(j>=vo[i]){
                    dp[i][j]=max(dp[i-1][j],dp[i-1][j-vo[i]]+va[i]);
                }
                else{
                    dp[i][j]=dp[i-1][j];
                }

                //cout<<i<<"  "<<j<<" "<<dp[i][j]<<endl;
            }
        }
        printf("%d
",dp[n][v]);
    }
    return 0;
}