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

一个简单背包, 根据状态转移方程就能做了:

dp[j] = max(dp[j], dp[j-w[i]] + v[i]);

#include <bits/stdc++.h>
using namespace std;

int main() {
    int T;
    int n, v;
    int value[10005];
    int weight[10005];
    scanf("%d", &T);
    while(T--) {
        int dp[10005];
        memset(dp, 0, sizeof dp);
        scanf("%d%d", &n, &v);
        for(int i = 0; i < n; i++) {
            scanf("%d", &value[i]);
        }
        for(int j = 0; j < n; j++) {
            scanf("%d", &weight[j]);
        }
        for(int i = 0; i < n; i++) {
            for (int j = v; j >= weight[i]; j--) {
                dp[j] = max(dp[j], dp[j - weight[i]] + value[i]);
            }
        }
        printf("%d\n", dp[v]);
    }

    return 0;
}