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1 Answer

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by Goeduhub's Expert (5.8k points)
 
Best answer

For Theory of 0/1 Knapsack Problem Using Dynamic Approach Click here

Java Program

import java.util.Scanner;

public class Knap1 

{

    static int max(int a, int b) 

    { 

        return (a > b)? a : b; 

    }

    static int knapSack(int W, int wt[], int val[], int n)

    {

        int i, w;

        int [][]K = new int[n+1][W+1];

        for (i = 0; i <= n; i++)

        {

            for (w = 0; w <= W; w++)

            {

                if (i==0 || w==0)

                    K[i][w] = 0;

                else if (wt[i-1] <= w)

                    K[i][w] = max(val[i-1] + K[i-1][w-wt[i-1]],  K[i-1][w]);

                else

                    K[i][w] = K[i-1][w];

            }

        } 

        return K[n][W];

    }

    public static void main(String args[])

    {

        Scanner sc = new Scanner(System.in);

        System.out.println("Enter the number of items: ");

        int n = sc.nextInt();

        System.out.println("Enter the items weights: ");

        int []wt = new int[n];

        for(int i=0; i<n; i++)

            wt[i] = sc.nextInt();

        System.out.println("Enter the items values: ");

        int []val = new int[n];

        for(int i=0; i<n; i++)

            val[i] = sc.nextInt();

        System.out.println("Enter the maximum capacity: ");

        int W = sc.nextInt();

        System.out.println("The maximum value that can be put in a knapsack of capacity W is: " + knapSack(W, wt, val, n));

        sc.close();

    }

Output

Output


For more VTU IV Sem DAA Lab Experiments Click here


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