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

using namespace std;
vector<pair<int,int> >a;
//dp table is full of zeros
int n,s,dp[1002][1002];
void ini(){
    for(int i=0;i<1002;i++)
        for(int j=0;j<1002;j++)
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
	//we obtain quantity of elements and size of knapsack
	//we get value of elements
	for(int i=0;i<n;i++)
	//we get size of elements
	for(int i=0;i<n;i++)
	//initialize dp table
	//print answer
	return 0;

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