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C++ Program to Solve the Fractional Knapsack Problem
In Fractional knapsack problem, a set of items are given, each with a weight and a value. We need to break items for maximizing the total value of knapsack and this can be done in greedy approach.
Algorithm
Begin Take an array of structure Item Declare value, weight, knapsack weight and density Calculate density=value/weight for each item Sorting the items array on the order of decreasing density We add values from the top of the array to total value until the bag is full, i.e; total value <= W End
Example Code
#include <iostream>
#include <bits/stdc++.h>
using namespace std;
typedef struct {
int v;
int w;
float d;
} Item;
void input(Item items[],int sizeOfItems) {
cout << "Enter total "<< sizeOfItems <<" item's values and weight" <<
endl;
for(int i = 0; i < sizeOfItems; i++) {
cout << "Enter "<< i+1 << " V ";
cin >> items[i].v;
cout << "Enter "<< i+1 << " W ";
cin >> items[i].w;
}
}
void display(Item items[], int sizeOfItems) {
int i;
cout << "values: ";
for(i = 0; i < sizeOfItems; i++) {
cout << items[i].v << "\t";
}
cout << endl << "weight: ";
for (i = 0; i < sizeOfItems; i++) {
cout << items[i].w << "\t";
}
cout << endl;
}
bool compare(Item i1, Item i2) {
return (i1.d > i2.d);
}
float knapsack(Item items[], int sizeOfItems, int W) {
int i, j;
float totalValue = 0, totalWeight = 0;
for (i = 0; i < sizeOfItems; i++) {
items[i].d = (float)items[i].v / items[i].w; //typecasting done (v is int and w is also int so we get final value of d as int)
}
sort(items, items+sizeOfItems, compare);
/*
uncomment if u need to check the data after sortingis done
cout << "values : ";
for(i = 0; i < sizeOfItems; i++) {
cout << items[i].v << "\t";
}
cout << endl << "weights: ";
for (i = 0; i < sizeOfItems; i++) {
cout << items[i].w << "\t";
}
cout << endl << "ratio : ";
for (i = 0; i < sizeOfItems; i++) {
cout << items[i].d << "\t";
}
cout << endl;
*/
for(i=0; i<sizeOfItems; i++) {
if(totalWeight + items[i].w<= W) {
totalValue += items[i].v ;
totalWeight += items[i].w;
} else {
int wt = W-totalWeight;
totalValue += (wt * items[i].d);
totalWeight += wt;
break;
}
}
cout << "Total weight in bag " << totalWeight<<endl;
return totalValue;
}
int main() {
int W;
Item items[4];
input(items, 4);
cout << "Entered data \n";
display(items,4);
cout<< "Enter Knapsack weight \n";
cin >> W;
float mxVal = knapsack(items, 4, W);
cout << "Max value for "<< W <<" weight is "<< mxVal;
}Output
Enter total 4 item's values and weight Enter 1 V Enter 1 W Enter 2 V Enter 2 W Enter 3 V Enter 3 W Enter 4 V Enter 4 W Entered data values: 2 0 0 0 weight: 0 490642064 0 4196544 Enter Knapsack weight Total weight in bag 0 Max value for 0 weight is 2
Updated on: 17-Feb-2022
5K+ Views
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