Code:
// A Dynamic Programming based solution for 0-1 Knapsack problem
#include
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;
int K[n + 1][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()
{
cout << "Enter the number of items in a Knapsack:";
int n, W;
cin >> n;
int val[n], wt[n];
for (int i = 0; i < n; i++)
{
cout << "Enter value and weight for item " << i << ":";
cin >> val[i];
cin >> wt[i];
}
// int val[] = { 60, 100, 120 };
// int wt[] = { 10, 20, 30 };
// int W = 50;
cout << "Enter the capacity of knapsack";
cin >> W;
cout << knapSack(W, wt, val, n);
return 0;
}
Output:
Enter the number of items in a Knapsack:5
Enter value and weight for item 0:11 111
Enter value and weight for item 1:22 121
Enter value and weight for item 2:33 131
Enter value and weight for item 3:44 141
Enter value and weight for item 4:55 151
Enter the capacity of knapsack 300
99
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