Check out example codes for "knapsack". It will help you in understanding the concepts better.

Code Example 1

``````/* A Naive recursive implementation of 0-1 Knapsack problem */
class Knapsack
{

// A utility function that returns maximum of two integers
static int max(int a, int b) { return (a > b)? a : b; }

// Returns the maximum value that can be put in a knapsack of capacity W
static int knapSack(int W, int wt[], int val[], int n)
{
// Base Case
if (n == 0 || W == 0)
return 0;

// If weight of the nth item is more than Knapsack capacity W, then
// this item cannot be included in the optimal solution
if (wt[n-1] > W)
return knapSack(W, wt, val, n-1);

// Return the maximum of two cases:
// (1) nth item included
// (2) not included
else return max( val[n-1] + knapSack(W-wt[n-1], wt, val, n-1),
knapSack(W, wt, val, n-1)
);
}

// Driver program to test above function
public static void main(String args[])
{
int val[] = new int[]{60, 100, 120};
int wt[] = new int[]{10, 20, 30};
int  W = 50;
int n = val.length;
System.out.println(knapSack(W, wt, val, n));
}
}
/*This code is contributed by Rajat Mishra */``````

Code Example 2

``````#include<bits/stdc++.h>
using namespace std;
vector<pair<int,int> >a;
//dp table is full of zeros
int n,s,dp;
void ini(){
for(int i=0;i<1002;i++)
for(int j=0;j<1002;j++)
dp[i][j]=-1;
}
int f(int x,int b){
//base solution
if(x>=n or b<=0)return 0;
//if we calculate this before, we just return the answer (value diferente of 0)
if(dp[x][b]!=-1)return dp[x][b];
//calculate de answer for x (position) and b(empty space in knapsack)
//we get max between take it or not and element, this gonna calculate all the
//posible combinations, with dp we won't calculate what is already calculated.
return dp[x][b]=max(f(x+1,b),b-a[x].second>=0?f(x+1,b-a[x].second)+a[x].first:INT_MIN);
}
int main(){
//fast scan and print
ios_base::sync_with_stdio(0);cin.tie(0);
//we obtain quantity of elements and size of knapsack
cin>>n>>s;
a.resize(n);
//we get value of elements
for(int i=0;i<n;i++)
cin>>a[i].first;
//we get size of elements
for(int i=0;i<n;i++)
cin>>a[i].second;
//initialize dp table
ini();