动态编程背包K精确项目 [英] Dynamic Programming Knapsack K-exact items

问题描述

我发现了这个非常方便的示例代码，该代码实现了背包问题的DP解决方案（对发帖人表示敬意）。

I found this very handy example code which implements a DP solution to the knapsack problem (kudos to the person who posted it).

https://codereview.stackexchange.com/questions/20569/dynamic-programming-solution-to-knapsack-问题

我正在尝试对其进行修改，以包括对背包中k项数量的限制。

I am trying to modify it to include a constraint on the number of items k in the knapsack.

我添加了第三个参数

def knapsack(items, maxweight, maxitems):

，并对重构进行了如下修改：

and modified the reconstruction as follows:

while i > 0:

    if bestvalues[i][j] != bestvalues[i - 1][j] and len(reconstruction) < maxitems:
        reconstruction.append(items[i - 1])
        j -= items[i - 1][1]

    i -= 1

只要我输入足够的项目以进行选择，它将始终收敛到所需的k个项目。但是，我可以肯定地说，这并没有找到全局最优值的最近似值。经过一些搜索之后，我所读到的讨论都涉及添加第三个维度k并在重建之前考虑约束条件（我*认为这是在最佳价值评估期间）。

Provided I input enough items to choose from this will always converge to the desired k number of items. However, I am fairly certain that this is not finding the closest approximation of the global optimum. The discussions I have read after some searching refer to adding a third dimension k and accounting for the constraint before the reconstruction (I *think this would be during the best value assessment).

有人可以提供如何执行此操作的示例吗？理想情况下，一个工作正常的python示例将是很棒的，但我将接受伪代码。我已经读过一些使用符号的说明，但是我仍然不确定如何用k进行约束（除了我在这里所做的以外）。

Can someone provide an example of how to do this? Ideally a working python example would be fantastic but I'll settle for pseudocode. I have read a few instructions using notation but I am still not sure how to constrain with k (outside of what I have done here).

谢谢！

推荐答案

正如我在上面的注释中所述，第三个维度是必需的，我已经编写了一个递归动态编程解决方案：

As i stated in the comment above a third dimension is required, i have written a recursive dynamic programming solution :

#include<bits/stdc++.h>

using namespace std;

int noOfItems, items[100], maxWeight, maxItems, value[100];
int dp[100][1000][100];

int solve(int idx, int currentWeight, int itemsLeft){
    if(idx == noOfItems || itemsLeft == 0) return 0;
    if(dp[idx][currentWeight][itemsLeft] != -1) return dp[idx][currentWeight][itemsLeft];
    int v1 = 0, v2 = 0;
    //try to included the current item
    if(currentWeight >= items[idx]) v1 = solve(idx+1, currentWeight-items[idx], itemsLeft-1) + value[idx];
    //exclude current item
    v2 = solve(idx+1, currentWeight, itemsLeft);
    return dp[idx][currentWeight][itemsLeft] = max(v1, v2);
}

//print the contents of the knapsack
void print(int idx, int currentWeight, int itemsLeft){
    if(idx == noOfItems || itemsLeft == 0) return;
    int v1 = 0, v2 = 0;
    if(currentWeight >= items[idx]) v1 = solve(idx+1, currentWeight-items[idx], itemsLeft-1) + value[idx];
    v2 = solve(idx+1, currentWeight, itemsLeft);
    if(v1 >= v2){
        cout << idx << " " << items[idx] << " " << value[idx] << endl;
        print(idx+1, currentWeight-items[idx], itemsLeft-1);
        return;
    }else{
        print(idx+1, currentWeight, itemsLeft);
        return;
    }
}

int main(){
    cin >> noOfItems >> maxWeight >> maxItems;
    for(int i = 0;i < noOfItems;i++) cin >> items[i] >> value[i];
    memset(dp, -1, sizeof dp);
    cout << solve(0, maxWeight, maxItems) << endl;  //prints the maximum    value that we can get from the constraints
    cout << "Printing the elements in the knapsack" << endl;
    print(0, maxWeight, maxItems);
return 0;
}

到ideone的解决方案链接： https://ideone.com/wKzqXk

Link to solution on ideone : https://ideone.com/wKzqXk

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