knapsack 01

// A dynamic programming based
// solution for 0-1 Knapsack problem
#include <bits/stdc++.h>
using namespace std;
// 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)
{	int i, w;
	vector<vector<int>> K(n + 1, vector<int>(W + 1));
	// Build table K[][] in bottom up manner
	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 val[] = { 60, 100, 120 };
	int wt[] = { 10, 20, 30 };
	int W = 50;
	int n = sizeof(val) / sizeof(val[0]);
	cout << knapSack(W, wt, val, n);
	return 0;
}