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/* A Naive recursive implementation of | ||
0-1 Knapsack problem */ | ||
#include <bits/stdc++.h> | ||
using namespace std; | ||
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// A utility function that returns | ||
// maximum of two integers | ||
int max(int a, int b) { return (a > b) ? a : b; } | ||
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// Returns the maximum value that | ||
// can be put in a knapsack of capacity W | ||
int knapSack(int W, int wt[], int val[], int n) | ||
{ | ||
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// Base Case | ||
if (n == 0 || W == 0) | ||
return 0; | ||
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// 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); | ||
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// 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)); | ||
} | ||
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// Driver code | ||
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; | ||
} | ||
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// This code is contributed by rathbhupendra |