Unbounded Fractional Knapsack
Given weights and values of n items, the task is to put these items in a knapsack of capacity W to get the maximum total value in the knapsack, we can repeatedly put the same item and we can also put fraction of an item.
Examples:
Input: val[] = {14, 27, 44, 19}, wt[] = {6, 7, 9, 8}, W = 50
Output: 244.444Input: val[] = {100, 60, 120}, wt[] = {20, 10, 30}, W = 50
Output: 300
Approach: The idea here is to just find the item which has largest value to weight ratio. Then fill the whole knapsack with this item only in order to maximize the final value of the knapsack.
Below is the implementataion of the above approach:
C++
// C++ implementation of the approach #include <bits/stdc++.h> using namespace std; // Function to return the maximum required value float knapSack(int W, float wt[], float val[], int n) { // maxratio will store the maximum value to weight // ratio we can have for any item and maxindex // will store the index of that element float maxratio = INT_MIN; int maxindex = 0; // Find the maximum ratio for (int i = 0; i < n; i++) { if ((val[i] / wt[i]) > maxratio) { maxratio = (val[i] / wt[i]); maxindex = i; } } // The item with the maximum value to // weight ratio will be put into // the knapsack repeatedly until full return (W * maxratio); } // Driver code int main() { float val[] = { 14, 27, 44, 19 }; float wt[] = { 6, 7, 9, 8 }; int n = sizeof(val) / sizeof(val[0]); int W = 50; cout << knapSack(W, wt, val, n); return 0; } |
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Java
// Java implementation of the approach class GFG { // Function to return the maximum required value static float knapSack(int W, float wt[], float val[], int n) { // maxratio will store the maximum value to weight // ratio we can have for any item and maxindex // will store the index of that element float maxratio = Integer.MIN_VALUE; int maxindex = 0; // Find the maximum ratio for (int i = 0; i < n; i++) { if ((val[i] / wt[i]) > maxratio) { maxratio = (val[i] / wt[i]); maxindex = i; } } // The item with the maximum value to // weight ratio will be put into // the knapsack repeatedly until full return (W * maxratio); } // Driver code public static void main(String[] args) { float val[] = {14, 27, 44, 19}; float wt[] = {6, 7, 9, 8}; int n = val.length; int W = 50; System.out.println(knapSack(W, wt, val, n)); } } // This code is contributed by 29AjayKumar |
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Python3
# Python implementation of the approach import sys # Function to return the maximum required value def knapSack(W, wt, val, n): # maxratio will store the maximum value to weight # ratio we can have for any item and maxindex # will store the index of that element maxratio = -sys.maxsize-1; maxindex = 0; # Find the maximum ratio for i in range(n): if ((val[i] / wt[i]) > maxratio): maxratio = (val[i] / wt[i]); maxindex = i; # The item with the maximum value to # weight ratio will be put into # the knapsack repeatedly until full return (W * maxratio); # Driver code val = [ 14, 27, 44, 19 ]; wt = [ 6, 7, 9, 8 ]; n = len(val); W = 50; print(knapSack(W, wt, val, n)); # This code is contributed by Rajput-Ji |
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C#
// C# implementation of the approach using System; class GFG { // Function to return the maximum required value static float knapSack(int W, float []wt, float []val, int n) { // maxratio will store the maximum value to weight // ratio we can have for any item and maxindex // will store the index of that element float maxratio = int.MinValue; int maxindex = 0; // Find the maximum ratio for (int i = 0; i < n; i++) { if ((val[i] / wt[i]) > maxratio) { maxratio = (val[i] / wt[i]); maxindex = i; } } // The item with the maximum value to // weight ratio will be put into // the knapsack repeatedly until full return (W * maxratio); } // Driver code public static void Main() { float []val = {14, 27, 44, 19}; float []wt = {6, 7, 9, 8}; int n = val.Length; int W = 50; Console.WriteLine(knapSack(W, wt, val, n)); } } // This code is contributed by AnkitRai01 |
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Output:
244.444
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