2021-09-18 01:02:57 +08:00
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package DynamicProgramming;
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/* A Naive recursive implementation
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of 0-1 Knapsack problem */
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public class BruteForceKnapsack {
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2021-09-18 01:03:36 +08:00
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// A utility function that returns
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// maximum of two integers
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static int max(int a, int b) {
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return (a > b) ? a : b;
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}
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2021-09-18 01:02:57 +08:00
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2021-09-18 01:03:36 +08:00
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// Returns the maximum value that
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// can be put in a knapsack of
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// capacity W
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static int knapSack(int W, int wt[], int val[], int n) {
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// Base Case
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if (n == 0 || W == 0) return 0;
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2021-09-18 01:02:57 +08:00
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2021-09-18 01:03:36 +08:00
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// If weight of the nth item is
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// more than Knapsack capacity W,
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// then this item cannot be included
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// in the optimal solution
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if (wt[n - 1] > W) return knapSack(W, wt, val, n - 1);
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2021-09-18 01:02:57 +08:00
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2021-09-18 01:03:36 +08:00
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// Return the maximum of two cases:
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// (1) nth item included
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// (2) not included
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else
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return max(val[n - 1] + knapSack(W - wt[n - 1], wt, val, n - 1), knapSack(W, wt, val, n - 1));
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}
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2021-09-18 01:02:57 +08:00
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2021-09-18 01:03:36 +08:00
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// Driver code
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public static void main(String args[]) {
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int val[] = new int[] {60, 100, 120};
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int wt[] = new int[] {10, 20, 30};
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int W = 50;
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int n = val.length;
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System.out.println(knapSack(W, wt, val, n));
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}
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2021-09-18 01:02:57 +08:00
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}
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