NYOJ B. Bone Collector

B. 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

15 101 2 3 4 55 4 3 2 1

Sample Output

14

这道题和前面那一道苹果是差不多的。

代码：

#include<stdio.h>#include<string.h>#include<math.h>#include<algorithm>using namespace std;int f[1005]= {0},c[1005],w[1005];int main(){    int t;    scanf("%d",&t);    while(t--)    {        int n,v;        scanf("%d%d",&n,&v);        for(int i=0; i<n; i++)            scanf("%d",&w[i]);        for(int i=0; i<n; i++)            scanf("%d",&c[i]);        memset(f,0,sizeof(f));        for(int i=0; i<n; i++)        {            for(int j=v; j>=0; j--)            {                if(j>=c[i])                    f[j]=max(f[j],f[j-c[i]]+w[i]);            }        }        printf("%d\n",f[v]);    }    return 0;}