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更新：根据 rrenaud 的精彩评论，我想澄清一下，虽然我正在生成所有排列顺序并不重要（所以它不完全像一个旅行推销员问题......但仍然有点相似）。我专注于我的问题的观察结果（尽管欢迎任何关于如何加快速度的见解），结果是相互建立的，也许有一些方法（动态编程？）我可以用来建立以前的结果而不是重复大量的计算。查看这些结果，您可以看到结果只是相互构建（这样做是为了将项目数更改为 3 并在 for 循环中将最大值更改为 3）：

Grapes = 0
Strawberries = 0
Raspberries = 0
****
Grapes = 1
Strawberries = 0
Raspberries = 0
****
Grapes = 2
Strawberries = 0
Raspberries = 0
****
Grapes = 3
Strawberries = 0
Raspberries = 0
****
Grapes = 0
Strawberries = 1
Raspberries = 0
****
Grapes = 1
Strawberries = 1
Raspberries = 0
****
Grapes = 2
Strawberries = 1
Raspberries = 0
****
Grapes = 3
Strawberries = 1
Raspberries = 0
****
Grapes = 0
Strawberries = 2
Raspberries = 0
****
Grapes = 1
Strawberries = 2
Raspberries = 0
****
Grapes = 2
Strawberries = 2
Raspberries = 0
****
Grapes = 3
Strawberries = 2
Raspberries = 0
****
Grapes = 0
Strawberries = 3
Raspberries = 0
****
Grapes = 1
Strawberries = 3
Raspberries = 0
****
Grapes = 2
Strawberries = 3
Raspberries = 0
****
Grapes = 3
Strawberries = 3
Raspberries = 0
****
Grapes = 0
Strawberries = 0
Raspberries = 1
****
Grapes = 1
Strawberries = 0
Raspberries = 1
****
Grapes = 2
Strawberries = 0
Raspberries = 1
****
Grapes = 3
Strawberries = 0
Raspberries = 1
****
Grapes = 0
Strawberries = 1
Raspberries = 1
****
Grapes = 1
Strawberries = 1
Raspberries = 1
****
Grapes = 2
Strawberries = 1
Raspberries = 1
****
Grapes = 3
Strawberries = 1
Raspberries = 1
****
Grapes = 0
Strawberries = 2
Raspberries = 1
****
Grapes = 1
Strawberries = 2
Raspberries = 1
****
Grapes = 2
Strawberries = 2
Raspberries = 1
****
Grapes = 3
Strawberries = 2
Raspberries = 1
****
Grapes = 0
Strawberries = 3
Raspberries = 1
****
Grapes = 1
Strawberries = 3
Raspberries = 1
****
Grapes = 2
Strawberries = 3
Raspberries = 1
****
Grapes = 3
Strawberries = 3
Raspberries = 1
****
Grapes = 0
Strawberries = 0
Raspberries = 2
****
Grapes = 1
Strawberries = 0
Raspberries = 2
****
Grapes = 2
Strawberries = 0
Raspberries = 2
****
Grapes = 3
Strawberries = 0
Raspberries = 2
****
Grapes = 0
Strawberries = 1
Raspberries = 2
****
Grapes = 1
Strawberries = 1
Raspberries = 2
****
Grapes = 2
Strawberries = 1
Raspberries = 2
****
Grapes = 3
Strawberries = 1
Raspberries = 2
****
Grapes = 0
Strawberries = 2
Raspberries = 2
****
Grapes = 1
Strawberries = 2
Raspberries = 2
****
Grapes = 2
Strawberries = 2
Raspberries = 2
****
Grapes = 3
Strawberries = 2
Raspberries = 2
****
Grapes = 0
Strawberries = 3
Raspberries = 2
****
Grapes = 1
Strawberries = 3
Raspberries = 2
****
Grapes = 2
Strawberries = 3
Raspberries = 2
****
Grapes = 3
Strawberries = 3
Raspberries = 2
****
Grapes = 0
Strawberries = 0
Raspberries = 3
****
Grapes = 1
Strawberries = 0
Raspberries = 3
****
Grapes = 2
Strawberries = 0
Raspberries = 3
****
Grapes = 3
Strawberries = 0
Raspberries = 3
****
Grapes = 0
Strawberries = 1
Raspberries = 3
****
Grapes = 1
Strawberries = 1
Raspberries = 3
****
Grapes = 2
Strawberries = 1
Raspberries = 3
****
Grapes = 3
Strawberries = 1
Raspberries = 3
****
Grapes = 0
Strawberries = 2
Raspberries = 3
****
Grapes = 1
Strawberries = 2
Raspberries = 3
****
Grapes = 2
Strawberries = 2
Raspberries = 3
****
Grapes = 3
Strawberries = 2
Raspberries = 3
****
Grapes = 0
Strawberries = 3
Raspberries = 3
****
Grapes = 1
Strawberries = 3
Raspberries = 3
****
Grapes = 2
Strawberries = 3
Raspberries = 3
****
Grapes = 3
Strawberries = 3
Raspberries = 3
****

我还在学习算法，所以我的知识库有点有限。我基本上有一个递归，随着我向它添加更多输入，它会保持指数增长，但我想知道是否有什么可以用来让它不这样做的。

这是一个递归示例（java代码），它基本上采用一个项目列表并使用0-n数量计算出每个项目的所有组合（为简单起见，我的代码有2个项目，每个项目的数量为0到5。 .variables 可以在循环中更改）。为了使这更有趣，我有一个名为 Q 的变量，它进行一些随机处理并根据列表检查其值，然后才继续。我尽可能地评论了代码，希望它对你来说很容易阅读：

import java.util.Arrays;
import java.util.List;
//learning playground safe to delete

public class main {

    public static void main(String[] args) {
        System.out.println("Starting..");
        Integer number_of_items = 2; //how many items should we test with, over 7 or 8 takes a long time MAX is 11(based on the number of names we have below)
        long startTime = System.currentTimeMillis(); //start timer
        Integer[] integers_temp = new Integer[number_of_items]; // create a list with exactly the number of items specified above
        Arrays.fill(integers_temp, 0); // populate list with zeros 
        List<Integer> list_to_start = Arrays.asList(integers_temp); //set it as a list
        String[] name_of_list_to_start = new String[] {"Grapes", "Strawberries", "Raspberries", "Blackberries", "Pineapples", "Oranges", "Prunes", "Pears", "cherries", "Peaches", "Apples"};       
        List<Integer> numbers_to_choose_from = Arrays.asList(new Integer[] {0, 1,2,3,4,5,6,7,8,9,10}); //list of numbers program can choose from(could be anything,just learning)
        counter(list_to_start.size(), list_to_start, name_of_list_to_start, numbers_to_choose_from);
        long endTime = System.currentTimeMillis(); 
        System.out.println("Total execution time: " + (endTime-startTime));
    }

    private static void counter(int length, List<Integer> list_to_start, String[] name_of_list_to_start, List<Integer> numbers_to_choose_from) {
        // If we've gone through everything then return the results
        if (length == 0) {
            for (int i = 0; i<list_to_start.size(); i++) {
                System.out.println(name_of_list_to_start[i] + " = " + list_to_start.get(i));
            }
            System.out.println("****");
            return;
        }   
        //This part basically increments list_to_start and then the above part displays it.
        for (int i = 0; i<=5; i++) {
            int q = i +2;  //do anything here..random just for example, right now just takes the number in the loop and adds by 10
            //System.out.println(q);  // this area is looped as many times as i^items, yet seems like after the first run work is duplicated.
            if (length != 0 && numbers_to_choose_from.contains(q)) {
                list_to_start.set((length-1), q);
                counter((length-1), list_to_start, name_of_list_to_start, numbers_to_choose_from);
                list_to_start.set((length-1), 0);
            }
        }
    }
}

如果将数量更改为 0-10 并将项目更改为 10，则它变为（10^10 即 10000000000 循环）。当我分析代码时，它说大部分时间都花在迭代器上，这是有道理的，所以我需要以某种方式减少它花费在迭代上的时间（希望有一种方法可以让它每个项目迭代一次，而不是每次递归） .

我的观察是最初的 for 循环似乎是必要的，因为它根据列表计算和检查值，但是一旦完成，它循环的所有其他时间都会重复相同的过程。我对所有算法都非常陌生（但目前正在阅读算法介绍），所以我想知道是否有一种方法可以使这个没有指数，以便它可以处理更多变量（这是我程序的一部分，它要大得多, 但理想情况下, 我需要处理尽可能多但至少 0-100 数量的 100 件物品并在一小时内完成或其他工作失败. 我无法用 100^100 来实现这一点, 所以我为想法而奋斗）。

如果有人有任何提示，或者如果他们看到任何解决类似问题的算法，那就太好了，因为我可以研究他们使用的逻辑，看看它是否适用。如果有人有直接的解决方案，那就太好了！

我希望这个问题不会太长并且有意义（我试图尽可能详细）

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2 回答 2

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If you want generate all possible permutations, this belongs to EXPTIME-complete class like towers of Hanoi, its more difficult than NP-complete problems (because all of them belong to PSPACE), you can't expect faster than 100^100. But may be there is some heuristics for original problem, So it's better to say it. Also if your original problem is this, you can say your boss: No one in the world can do this. Also you can make it parallel, but except using extremely powerful mainframe, you don't have any chance to get result.

于 2012-04-06T18:51:06.433 回答
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现在的代码正在迭代每个组合。对于 n=100，没有办法在合理的时间内运行。优化内部循环不会阻止它呈指数增长。

您将需要一种不同的方法来使其可行。如何做到这一点取决于您要解决的问题的详细信息。例如，一些优化问题可以通过动态规划、线性规划、最大流量或二分搜索来解决。其他问题通过寻找近似解来解决。

于 2012-04-06T19:01:04.497 回答