Modifed version of the knapsack - c++

Question

I know how to code the solution for the 0-1 knapsack problem. However, I do not know how to change this code such that we can pick multiple copies of the same item. So, for each item i , we have another paramter k(i) which denotes the number of copies of i present. I will be obliged if someone could help me with this. Below is my code when k(i) = 1. Look at the knapsack function for a top down dp solution
#include<cstring>
#include<cstdio>
#include<iostream>
using namespace std;
int n, g;
int p[1005],w[1005];
int dp[1004][35];// num of object and weight left
int knapsack(int resourcel, int item){
//if(resourcel < 0) return (1<<31);
if(item == n) return 0;
if(resourcel == 0) return 0;
if(dp[item][resourcel] != -1) return dp[item][resourcel];
if(resourcel - w[item] < 0){
dp[item][resourcel] = knapsack(resourcel,item+1);
return dp[item][resourcel];
}
else{
int take = knapsack(resourcel - w[item],item+1) + p[item];
int notTake = knapsack(resourcel,item+1);
dp[item][resourcel] = take > notTake?take : notTake;
return dp[item][resourcel];
}
}
int main(){
int tc,dummy, sum =0;
//freopen("in.txt","r",stdin);
scanf("%d",&tc);
for(int i = 0 ; i < tc; i++){
sum = 0;
memset(dp,-1,sizeof(dp));
scanf("%d",&n);
//cout<<" n is : "<<n<<endl;
for(int j = 0 ; j < n ;j++){
scanf("%d %d",&p[j],&w[j]);
//cout<<" price and val is : "<<p[j]<<" " << w[j]<<endl;
}
scanf("%d",&g);
//cout<<"g is : "<<g<<endl;
for(int p = 0 ; p< g;p++){
scanf("%d",&dummy);
sum+= knapsack(dummy,0);//wight allowed and item visited
}
printf("%d\n",sum);
}
return 0;
}

Answer 1

Your knapsack code is overly complicated. Here is another way to do it:
Let dp[i] = maximum profit we can get for weight i.
for i = 1 to numItems do
for j = knapsackWeight down to items[i].weight do
dp[j] = max(dp[j], dp[j - items[i].weight] + items[i].profit)
Now, you also want a field item.copies. We can simply add another loop in the middle to iterate this many times.
for i = 1 to numItems do
for k = 1 to items[i].copies do
for j = knapsackWeight down to items[i].weight do
dp[j] = max(dp[j], dp[j - items[i].weight] + items[i].profit)