背包問(wèn)題 小灰

Prerequisites: Algorithm for fractional knapsack problem

先決條件： 分?jǐn)?shù)背包問(wèn)題算法

Here, we are discussing the practical implementation of the fractional knapsack problem. It can be solved using the greedy approach and in fractional knapsack problem, we can break items i.e we can take a fraction of an item. For examples, you can read this article first.

在這里，我們正在討論小背包問(wèn)題的實(shí)際實(shí)現(xiàn) 。 可以使用貪婪方法解決該問(wèn)題，在小背包問(wèn)題中，我們可以破壞物品，即可以取物品的一小部分。 例如，您可以先閱讀本文 。

Problem statement: We have given items i₁, i₂, ..., i_n (item we want to put in our bag) with associated weights w₁, w₂, ..., w_n and profit values P₁ , P₂ ,..., P_n. Now problem is how we can maximize the total benefit given capacity of bag is C?

問(wèn)題陳述：我們給了項(xiàng)目i ₁ ， i ₂ ，...， i _n (我們要放入袋中的項(xiàng)目)，其權(quán)重為w ₁ ， w ₂ ，...， w _n和利潤(rùn)值P ₁ ， P ₂ ，...， P _n 。 現(xiàn)在的問(wèn)題是，在袋子容量為C的情況下， 如何才能使總收益最大化 ？

Algorithm:

算法：

Compute the profit per weight density for each item using the formula d_i = P_i / w_i.

使用公式d _i = P _i / w _i計(jì)算每個(gè)項(xiàng)目的每重量密度利潤(rùn)。

Sort each item by its profit per weight density.

按每重量密度的利潤(rùn)對(duì)每個(gè)項(xiàng)目進(jìn)行排序。

Maximize the profit i.e Take as much as possible of the profit per weight density item not already in the bag.

最大化利潤(rùn)，即盡可能多地利用袋中尚未存在的每重量密度物品的利潤(rùn)。

分?jǐn)?shù)背負(fù)問(wèn)題的C ++實(shí)現(xiàn) (C++ implementation of fractional knapsack problem)

#include <bits/stdc++.h> using namespace std; // Structure for an item struct myItem {int itemNo; int profit;int weight;float ppw; // profit per weight }; // Comparison function to sort Item according to profit per weight ratio bool cmp(struct myItem a, struct myItem b) { return a.ppw > b.ppw; } float fractionalKnapsack(int Capacity, struct myItem arr[], int n) { //calculating profit per weight ratiofor(int i=0;i<n;i++){arr[i].ppw = ((float)arr[i].profit / arr[i].weight);}// sorting Item on basis of profit per weight ratio sort(arr, arr + n, cmp); cout<<"details of all items : \n";cout<<"itemNo\t"<<"Profit\t"<<"Weight\t"<<"PPW\t"<<endl;for (int i = 0; i < n; i++) { cout <<arr[i].itemNo<<"\t"<<arr[i].profit<<"\t"<<arr[i].weight<<"\t"<<((float)arr[i].ppw)<<endl; }cout<<endl;float Max = 0.0; // Maximum profitint i=0; //take items until capacity becomes zerowhile(Capacity > 0 && i<=n-1){// if we can take all weights of itemif(Capacity >= arr[i].weight){Max = Max + (float)arr[i].profit;Capacity = Capacity - arr[i].weight;}// we can take only fraction of itemelse{Max += (Capacity * arr[i].ppw);Capacity = 0;}i++;} return Max; } // driver function int main() { int C = 25; // Capacity of knapsack myItem arr[] = { {1, 30, 10, 0}, {2, 20, 5, 0} , {3, 40, 15, 0}, {4, 36, 8, 0}}; int n = sizeof(arr) / sizeof(arr[0]);cout<<"details of all items before sorting and without calculating PPW: \n";cout<<"itemNo\t"<<"Profit\t"<<"Weight\t"<<"PPW\t"<<endl;for (int i = 0; i < n; i++) { cout <<arr[i].itemNo<<"\t"<<arr[i].profit<<"\t"<<arr[i].weight<<"\t"<<((float)arr[i].ppw)<<endl; } cout<<endl;cout << "Maximum profit we can obtain = "<< fractionalKnapsack(C, arr, n);return 0; }

Output

輸出量

details of all items before sorting and without calculating PPW: itemNo Profit Weight PPW 1 30 10 0 2 20 5 0 3 40 15 0 4 36 8 0details of all items : itemNo Profit Weight PPW 4 36 8 4.5 2 20 5 4 1 30 10 3 3 40 15 2.66667Maximum profit we can obtain = 91.3333

翻譯自: https://www.includehelp.com/icp/fractional-knapsack-problem.aspx

背包問(wèn)題 小灰

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