CQUOJ 2500 - 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 2³¹).

 

Sample Input

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

Sample Output

14

 

/*
2016年4月22日23:42:23
    01背包 

*/


# include <iostream>
# include <cstdio>
# include <cstring>
# include <algorithm>
# include <queue>
# include <vector>
# include <cmath>
# define INF 0x3f3f3f3f
using namespace std;
const int N = 1003;

int main(void)
{
    int test, i, j, n, m;
    int v[N], w[N], dp[N];
    scanf("%d", &test);
    while (test--){
        scanf("%d %d", &n, &m);
        for (i = 1; i <= n; i++)
            scanf("%d", &v[i]);
        for (i = 1; i <= n; i++)
            scanf("%d", &w[i]);
        memset(dp, 0, sizeof(dp));
        dp[0] = 0;
        for (i = 1; i <= n; i++){
            for (j = m; j >= w[i]; j--)
                dp[j] = max(dp[j], dp[j-w[i]]+v[i]);
        }
        printf("%d\n", dp[m]);
    } 
    return 0;    
}