Bone Collector hdu2602（經典dp）01揹包

Bone Collector

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 59894    Accepted Submission(s): 24996

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<cstdio>
#include<iostream>
#include<cmath>
#include<cstring>
#define mem(a,b) memset(a,b,sizeof(a))
using namespace std;
int v[100005],w[10005],dp[10005];
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        mem(v,0);
        mem(w,0);
        mem(dp,0);
        int n,m;
        scanf("%d%d",&n,&m);
        for(int i=1;i<=n;i++)
        {
            scanf("%d",&v[i]);
        }
         for(int i=1;i<=n;i++)
        {
            scanf("%d",&w[i]);
        }
        for(int i=1;i<=n;i++)
        {
            for(int j=m;j>=0;j--)
            {
                if(j>=w[i])
                    dp[j]=max(dp[j],dp[j-w[i]]+v[i]);
            }
        }
        printf("%d\n",dp[m]);
    }
    return 0;
}