hdu 2062

Bone Collector

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 13436    Accepted Submission(s): 5273

Problem 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 2³¹).

 

 

Sample Input

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

 

 

Sample Output

14

 #include<iostream>
using namespace std;
#define max(a,b) a>b?a:b
int a[1010],b[1010],c[1010][1010];
int main()
{
    int i,j,l,m,n;
    while(cin>>m)
    {
        while(m--)
        {
            cin>>n>>l;
            memset(a,0,sizeof(a));
            memset(b,0,sizeof(b));
            memset(c,0,sizeof(c));
            for(i=1;i<=n;i++)
                cin>>a[i];
            for(i=1;i<=n;i++)
                cin>>b[i];
            for(i=1;i<=n;i++)
            {
                for(j=0;j<=l;j++)
                {
                    if(j>=b[i])
                        c[i][j]=max(c[i-1][j],c[i-1][j-b[i]]+a[i]);
                    else
                        c[i][j]=c[i-1][j];
                }
            }
            cout<<c[n][l]<<endl;
        }
    }
    return 0;
}