Prerrequisito: Problema de mochila fraccional
los pesos N W máximo
Nota:
C++
// 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;
}
C++
// 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);
}
Publicación traducida automáticamente
Artículo escrito por ishujain18123 y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA