C++ Program for the Fractional Knapsack Problem

Pre-requisite: Fractional Knapsack Problem

Given two arrays weight[] and profit[] the weights and profit of N items, we need to put these items in a knapsack of capacity W to get the maximum total value in the knapsack.
Note: Unlike 0/1 knapsack, you are allowed to break the item.

Examples:

Input: weight[] = {10, 20, 30}, profit[] = {60, 100, 120}, N= 50
Output: Maximum profit earned = 240
Explanation: 
Decreasing p/w ratio[] = {6, 5, 4}
Taking up the weight values 10, 20, (2 / 3) * 30 
Profit = 60 + 100 + 120 * (2 / 3) = 240

Input: weight[] = {10, 40, 20, 24}, profit[] = {100, 280, 120, 120}, N = 60
Output: Maximum profit earned = 440
Explanation: 
Decreasing p/w ratio[] = {10, 7, 6, 5}
Taking up the weight values 10, 40, (1 / 2) * 120 
Profit = 100 + 280 + (1 / 2) * 120 = 440



 

Method 1 – without using STL: The idea is to use Greedy Approach. Below are the steps:

  1.  Find the ratio value/weight for each item and sort the item on the basis of this ratio.
  2. Choose the item with the highest ratio and add them until we can’t add the next item as a whole.
  3. In the end, add the next item as much as we can.
  4. Print the maximum profit after the above steps.

Below is the implementation of the above approach:
 

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// C++ program to solve fractional
// Knapsack Problem
#include <bits/stdc++.h>
  
using namespace std;
  
// Structure for an item which stores
// weight & corresponding value of Item
struct Item {
    int value, weight;
  
    // Constructor
    Item(int value, int weight)
        : value(value), weight(weight)
    {
    }
};
  
// Comparison function to sort Item
// according to val/weight ratio
bool cmp(struct Item a, struct Item b)
{
    double r1 = (double)a.value / a.weight;
    double r2 = (double)b.value / b.weight;
    return r1 > r2;
}
  
// Main greedy function to solve problem
double fractionalKnapsack(struct Item arr[],
                          int N, int size)
{
    // Sort Item on basis of ratio
    sort(arr, arr + size, cmp);
  
    // Current weight in knapsack
    int curWeight = 0;
  
    // Result (value in Knapsack)
    double finalvalue = 0.0;
  
    // Looping through all Items
    for (int i = 0; i < size; i++) {
  
        // If adding Item won't overflow,
        // add it completely
        if (curWeight + arr[i].weight <= N) {
            curWeight += arr[i].weight;
            finalvalue += arr[i].value;
        }
  
        // If we can't add current Item,
        // add fractional part of it
        else {
            int remain = N - curWeight;
            finalvalue += arr[i].value
                          * ((double)remain
                             / arr[i].weight);
  
            break;
        }
    }
  
    // Returning final value
    return finalvalue;
}
  
// Driver Code
int main()
{
    // Weight of knapsack
    int N = 60;
  
    // Given weights and values as a pairs
    Item arr[] = { { 100, 10 },
                   { 280, 40 },
                   { 120, 20 },
                   { 120, 24 } };
  
    int size = sizeof(arr) / sizeof(arr[0]);
  
    // Function Call
    cout << "Maximum profit earned = "
         << fractionalKnapsack(arr, N, size);
    return 0;
}

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Output:

Maximum profit earned = 440

Time Complexity: O(N*log2N)
Auxiliary Space: O(1)

Method 2 – using STL:

  1. Create a map with profit[i] / weight[i] as first and i as Second element for each element.
  2. Define a variable max_profit = 0.
  3. Traverse the map in reverse fashion:
    • Create a variable named fraction whose value is equivalent to remaining_weight / weight[i].
    • If remaining_weight is greater than or equals to zero and its value is greater than weight[i] add current profit to max_profit and reduce the remaining weight by weight[i].
    • Else if remaining weight is less than weight[i] add fraction * profit[i] to max_profit and break.
  4. Print the max_profit.

 Below is the implementation of the above approach:

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// C++ program to Fractional Knapsack
// Problem using STL
#include <bits/stdc++.h>
using namespace std;
  
// Function to find maximum profit
void maxProfit(vector<int> profit,
               vector<int> weight, int N)
{
  
    // Number of total weights present
    int numOfElements = profit.size();
    int i;
  
    // Multimap container to store
    // ratio and index
    multimap<double, int> ratio;
  
    // Variable to store maximum profit
    double max_profit = 0;
    for (i = 0; i < numOfElements; i++) {
  
        // Insert ratio profit[i] / weight[i]
        // and corresponding index
        ratio.insert(make_pair(
            (double)profit[i] / weight[i], i));
    }
  
    // Declare a reverse iterator
    // for Multimap
    multimap<double, int>::reverse_iterator it;
  
    // Traverse the map in reverse order
    for (it = ratio.rbegin(); it != ratio.rend();
         it++) {
  
        // Fraction of weight of i'th item
        // that can be kept in knapsack
        double fraction = (double)N / weight[it->second];
  
        // if remaining_weight is greater
        // than the weight of i'th item
        if (N >= 0
            && N >= weight[it->second]) {
  
            // increase max_profit by i'th
            // profit value
            max_profit += profit[it->second];
  
            // decrement knapsack to form
            // new remaining_weight
            N -= weight[it->second];
        }
  
        // remaining_weight less than
        // weight of i'th item
        else if (N < weight[it->second]) {
            max_profit += fraction
                          * profit[it->second];
            break;
        }
    }
  
    // Print the maximum profit earned
    cout << "Maximum profit earned is:"
         << max_profit;
}
  
// Driver Code
int main()
{
    // Size of list
    int size = 4;
  
    // Given profit and weight
    vector<int> profit(size), weight(size);
  
    // Profit of items
    profit[0] = 100, profit[1] = 280,
    profit[2] = 120, profit[3] = 120;
  
    // Weight of items
    weight[0] = 10, weight[1] = 40,
    weight[2] = 20, weight[3] = 24;
  
    // Capacity of knapsack
    int N = 60;
  
    // Function Call
    maxProfit(profit, weight, N);
}

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Output:

Maximum profit earned is:440

Time Complexity: O(N)
Auxiliary Space: O(N)

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