Bone Collector（hdu2602）-01背包

Bone Collector

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 2 ³¹).

Sample Input

1
5 10
1 2 3 4 5
5 4 3 2 1

Sample Output

14

#include<cstring>
#include<cstdio>
#include<iostream>
using namespace std;
const int maxn=1010;

int dp[maxn][maxn];//即dp[i][v]表示前i件物品恰放入一个容量为v的背包可以获得的最大价值。
int main()
{
    int n,m,t;
    scanf("%d",&t);
    while(t--){
        int v[maxn],w[maxn];
        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]);
        memset(dp,0,sizeof(dp));
        for(int i=1;i<=n;i++){
            for(int j=0;j<=m;j++){
                if(w[i]<=j)
                    dp[i][j]=max(dp[i-1][j],dp[i-1][j-w[i]]+v[i]);
                else
                    dp[i][j]=dp[i-1][j];
            }
        }
        printf("%d\n",dp[n][m]);
    }
    return 0;
}