0-1 Knapsack Problem in C Using Dynamic Programming

Here you will learn about 0-1 knapsack problem in C.

We are given n items with some weights and corresponding values and a knapsack of capacity W. The items should be placed in the knapsack in such a way that the total value is maximum and total weight should be less than knapsack capacity.

In this problem 0-1 means that we can’t put the items in fraction. Either put the complete item or ignore it. Below is the solution for this problem in C using dynamic programming.

Program for Knapsack Problem in C Using Dynamic Programming

#include<stdio.h>
 
int max(int a, int b) { return (a > b)? a : b; }
 
int knapSack(int W, int wt[], int val[], int n)
{
   int i, w;
   int K[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];
}
 
int main()
{
    int i, n, val[20], wt[20], W;
    
    printf("Enter number of items:");
    scanf("%d", &n);
    
    printf("Enter value and weight of items:\n");
    for(i = 0;i < n; ++i){
    	scanf("%d%d", &val[i], &wt[i]);
    }

    printf("Enter size of knapsack:");
    scanf("%d", &W);
    
    printf("%d", knapSack(W, wt, val, n));
    return 0;
}

Output

Enter number of items:3
Enter value and weight of items:
100 20
50 10
150 30
Enter size of knapsack:50
250

You can watch below video to learn knapsack problem easily.

Source: http://www.geeksforgeeks.org/dynamic-programming-set-10-0-1-knapsack-problem/

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