C++ 程序解决分数背包问题


在分数背包问题中,给定了一组物品,每个物品都具有重量和价值。我们需要打破物品以最大化背包的总价值,这可以通过贪婪算法来完成。

算法

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

示例代码

#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;
}

输出

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

更新于:17-2-2022

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