Notes

/* A Naive recursive implementation
of 0-1 Knapsack problem */
#include <stdio.h>

// A utility function that returns
// maximum of two integers
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
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
int main()
{
int val[] = { 60, 100, 120 };
int wt[] = { 10, 20, 30 };
int W = 50;
int n = sizeof(val) / sizeof(val[0]);
printf("%d", knapSack(W, wt, val, n));
return 0;
}