DP: Bone Collector

Bone Collector Time Limit:1000MS    Memory Limit:32768KB    64bit IO Format:%I64d & %I64u

Submit Status Practice HDU 2602

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 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 <cstring>
    #include <iostream>

    using namespace std;

    int main(){
        int N, V, v, i;
        long long val[1010], c[1010], ans[1010];
        int t;
        cin >> t;
        while(t --){
            cin >> N >> V;
            memset(val, 0, sizeof(val));
            memset(c, 0, sizeof(c));
            memset(ans, 0, sizeof(ans));
            for(i = 1; i <= N; i ++)
                scanf("%I64d", &val[i]);
            for(i = 1; i <= N; i ++)
                scanf("%I64d", &c[i]);
            for(i=1; i <= N; i ++){
                for(v = V; v >= c[i]; v--){
                   ans[v] = max (ans[v], ans[v-c[i]] + val[i]);
                }
            }
            cout << ans[V] << endl;
        }
        return 0;
    }