Bone Collector

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

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

简单01背包

#include<iostream>
#include<cstdio>
#include<cmath>
#include<queue>
#include<algorithm>
#include<cstring>
using namespace std;
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        int n,i,j,v;
        scanf("%d%d",&n,&v);
        int dp[1100]={0},c[1100],w[1100];
        for(i=0;i<n;i++)
            scanf("%d",w+i);
        for(i=0;i<n;i++)
            scanf("%d",c+i);
        for(i=0;i<n;i++)
        {
            for(j=v;j>=c[i];j--)
            {
                if(dp[j]<dp[j-c[i]]+w[i])
                {
                    dp[j]=dp[j-c[i]]+w[i];
                }
            }
        }
        printf("%d\n",dp[v]);
    }
    return 0;
}

HDU2602Bone Collector