0_1_Knapsack_problem.cpp

// A Dynamic Programming based
// solution for 0-1 Knapsack problem
#include <stdio.h>

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)
{
	int i, w;
	int K[n + 1][W + 1];

	// The state K[i][j] will denote the maximum value of ‘j-weight’ considering all values from ‘1 to ith‘.
	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];
}

// Driver Code
int main()
{
	int profit[] = { 60, 100, 120 };
	int weight[] = { 10, 20, 30 };
	int W = 50;
	int n = sizeof(profit) / sizeof(profit[0]);
	printf("%d", knapSack(W, weight, profit, n));
	return 0;
}