분수 배낭 문제를 해결하기 위한 C++ 프로그램

분수 배낭 문제에서는 각 항목에 가중치와 값이 있는 항목 집합이 제공됩니다. 배낭의 총 가치를 극대화하기 위해 아이템을 깨야 하는데 이것은 욕심부리는 방식으로 가능합니다.

알고리즘

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

예시 코드

#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;
}

출력

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