![](/img/trans.png)
[英]PYTHON : Find all combinations from a list with condition using recursion
[英]Python to C++: Algorithm that list all combinations of Knapsack using recursion
我正在尝试实现一个代码,该代码使用递归列出背包问题的所有可能组合。 我在递归上有困难。 我试图解决它但一无所获,所以我做了一些研究,并在
Java
Python 中找到了一个代码,但我很难尝试用 C++ 重写该代码。
这是
Java
Python中的解决方案代码:
items = [1,1,3,4,5]
knapsack = []
limit = 7
def print_solutions(current_item, knapsack, current_sum):
#if all items have been processed print the solution and return:
if current_item == len(items):
print knapsack
return
#don't take the current item and go check others
print_solutions(current_item + 1, list(knapsack), current_sum)
#take the current item if the value doesn't exceed the limit
if (current_sum + items[current_item] <= limit):
knapsack.append(items[current_item])
current_sum += items[current_item]
#current item taken go check others
print_solutions(current_item + 1, knapsack, current_sum )
print_solutions(0,knapsack,0)
我在此链接中找到了该代码
这是我试图做的..
#include <iostream>
using namespace std;
void AddItem(int item, int *knapsack) {
int i = 0;
while (knapsack[i] != -1)
i++;
knapsack[i] = item;
};
void printKnapsack(int *knapsack, int n) {
cout << "[";
for (int i = 0; i < n; i++)
if (knapsack[i] != -1)
cout << knapsack[i] << ",";
}
void print_solutions(int current_item, int *knapsack, int current_sum, int *items, int n, int limit) {
//if all items have been processed print the solution and return
if (current_item == n - 1) {
printKnapsack(knapsack, n);
return;
};
//don't take the current item and go check others
print_solutions(current_item + 1, knapsack, current_sum, items, n, limit);
//take the current item if the value doesn't exceed the limit
if (current_sum + items[current_item] <= limit) {
AddItem(items[current_item], knapsack);
current_sum += items[current_item];
};
//current item taken go check others
print_solutions(current_item + 1, knapsack, current_sum, items, n, limit);
};
int main() {
int current_item = 0;
int current_sum = 0;
int limit, n;
cout << "Type the maximum weight ";
cin >> limit;
cout << "How many items? ";
cin >> n;
int* knapsack;
knapsack = new int[10];
for (int i = 0; i < 10; i++)
knapsack[i] = -1;
int * items;
items = new int[n];
cout << "Type weights.";
for (int i = 0; i < n; i++) {
cin >> items[i];
};
print_solutions(0, knapsack, 0, items, n, limit);
return 0;
}
输入:
7 // limit
5 // number of items
1 1 3 4 5 // items
我希望得到以下最终结果:
[]
[5]
[4]
[3]
[3, 4]
[1]
[1, 5]
[1, 4]
[1, 3]
[1]
[1, 5]
[1, 4]
[1, 3]
[1, 1]
[1, 1, 5]
[1, 1, 4]
[1, 1, 3]
但我得到的只是用 3 和 4 填充的数组,而不是得到所有实际的解决方案。
在您将算法从 python 转录到 C++ 时,存在一个与参数传递相关的语言语义相关的主要问题。
在 python 中,您编写以下内容:
print_solutions(current_item + 1, list(knapsack), current_sum)
然后list(knapsack)
是来自knapsack
列表的副本。 所以中间的递归调用保持原来的knapsack
不变,而第二个递归调用改变了原来的knapsack
:
print_solutions(current_item + 1, knapsack, current_sum)
然而,在您的 C++ 代码中,在这两种情况下,您都在处理原始knapsack
列表(数组参数通过引用传递),因此knapsack
完全搞砸了:
//don't take the current item and go check others
print_solutions(current_item + 1, knapsack, current_sum, items, n, limit);
您可以创建一个临时数组并在其中复制knapsack
,或者更好的是,您开始使用vector
,这将使您的 C++ 生活更轻松(注意按值传递或按引用传递)。
以下版本使用向量。 参数中的&
表示它是通过引用传递的参数(即可以更改原始向量)。 请注意,我们不再需要传递n
,因为向量知道它的长度,就像 Python 中的 list 一样:
void print_solutions(int current_item, vector<int>& knapsack, int current_sum, const vector<int>& items, int limit) {
//if all items have been processed print the solution and return
if (current_item == items.size() ) {
printKnapsack(knapsack);
return;
};
//don't take the current item and go check others
vector<int> knapcopy = knapsack;
print_solutions(current_item + 1, knapcopy, current_sum, items, limit);
//take the current item if the value doesn't exceed the limit
if (current_sum + items[current_item] <= limit) {
knapsack.push_back(items[current_item]);
current_sum += items[current_item];
//current item taken go check others
print_solutions(current_item + 1, knapsack, current_sum, items, limit);
};
};
这里有一个在线演示。
声明:本站的技术帖子网页,遵循CC BY-SA 4.0协议,如果您需要转载,请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.