A - 01背包

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 
       
       

 

代码：

#include<stdio.h>
#include<string.h>
//#include<math.h>
#include<algorithm>
using namespace std;
/*struct bone
{
    int va,vo;
} a[1100];
bool cmp(bone x,bone y)
{
    return x.va>y.va;
}*/
int main()
{
    int t,n,v,i,j,h[1100],sum,vol[1100],val[1100];
    scanf("%d",&t);
    while(t--)
    {
        sum=0;
        scanf("%d %d",&n,&v);
        for(i=0; i<n; i++)
            scanf("%d",&val[i]);
        for(i=0; i<n; i++)
            scanf("%d",&vol[i]);
        //sort(a,a+n,cmp);
        memset(h,0,sizeof(h));
        for(i=0; i<n; i++)
        {
            for(j=v; j>=vol[i]; j--)
            {
                sum=h[j-vol[i]]+val[i];
                //if(h[j]<sum)
                    //h[j]=sum;
                    h[j]=max(h[j],sum);
            }
        }
        printf("%d\n",h[v]);


    }
    return 0;
}