/* A naive recursive implementation
* of 0-1 Knapsack problem
*/
#include <stdio.h>
// A utility function that returns
// maximum of two intergers
int max(int left, int right){
return left > right ? left: right;
}
/*
* Returns the maximum value that can be
* put in a knapsack of capacity W
*/
int knapsack(int W, int wt[], int vl[], int n){
//Base case
if(W == 0 || n == 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, vl, n - 1);
/**
* Return the maximum of two cases:
* (1) nth item included
* (2) not included
*/
else
return max(vl[n - 1] + knapsack(W - wt[n - 1], wt, vl, n-1),
knapsack(W, wt, vl, n - 1)
);
}
//Driver program to test above function
int main(){
int wt[] = {10, 20, 30};
int vl[] = {60, 100, 120};
int n = sizeof(wt) / sizeof(wt[0]);
printf("The maximum sum of values:%5d\n", knapsack(50, wt, vl,n));
return 0;
}