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

 

Sample Output

 

Author

Teddy

 

Source

 

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 #include<cstdio>

 #include<cstring>

 #include<iostream>

 #include<stack>

 #include<set>

 #include<map>

 #include<queue>

 #include<algorithm>

 using namespace std;

 #define max(a,b) (a>b?a:b)

 int va[],vo[],dp[][];

 int main()

 {

     //freopen("D:\\INPUT.txt","r",stdin);

     int t,i,j;

     int n,v;

     scanf("%d",&t);

     while(t--)

     {

         scanf("%d %d",&n,&v);

         for(i=; i<=n; i++)

         {

             scanf("%d",&va[i]);

         }

         for(i=; i<=n; i++)

         {

             scanf("%d",&vo[i]);

         }

         //dp[i][j]  前i件物品放入j体积的价值的最大值

         //dp[i][j]=max(dp[i-1][j],dp[i-1][j-vo[i]]+va[i])

         for(i=; i<=n; i++) //i体积

         {

             for(j=; j<=v; j++)

             {

                 if(j>=vo[i]){

                     dp[i][j]=max(dp[i-][j],dp[i-][j-vo[i]]+va[i]);

                 }

                 else{

                     dp[i][j]=dp[i-][j];

                 }

                 //cout<<i<<"  "<<j<<" "<<dp[i][j]<<endl;

             }

         }

         printf("%d\n",dp[n][v]);

     }

     return ;

 }