如何从一组k个元素中生成长度为n的所有排列 [英] How to generate all permutations of lenght n from a set of k elements

问题描述

例如，我将元素 [1,2,3,4,5] 设置为 k = 5 ，我希望所有长度为 n的排列= 2 .

For example I have this set k=5 of elements [1,2,3,4,5] and I want all permutations of length n=2.

1,2
1,3
1,4
1,5
2,1
etc etc. 

问题是我不能使用STL，外部数学库等.

Thing is i can't use STL, external math libraries etc.

我已经在这个问题上坐了大约3天了，我要疯了.我尝试的是使用Heap算法生成所有元素的所有排列，然后在所有k个排列的前n个数字中包含n个元素的所有排列，我可以截断并删除重复项，但是复杂度是太高了(n！)

I've been sitting on this problem for about 3 days now and im about to go crazy. What i tried is generating all permutations of all the elements using Heap's algorithm, and then all the permutations of n elements where contained in the first n numbers of all k-permutations and i could just truncate and delete duplicates, but then the complexity is way too high(n!)

我知道这个问题有一个很好的解决方案，因为我已经看到额外的模块/库可以解决有关置换字符串的问题

I know the problem has a good solution as I've seen this being done with extra modules/libraries in questions about permutating strings

更多信息:我只需要用它就可以强行解决不平衡的分配问题，而当我大声强力"解决问题时，匈牙利算法似乎太长了.我的方法没有接近允许的执行时间，因为当我有一个大小为8x3的数组时，我的算法需要8！比较绝对可以将其优化为小得多的数量.

extra info : I only need this to brute force an unbalanced assignment problem, and Hungarian algorithm seems way too long when i'm aloud to "brute-force" the problem. My approach didn't come close to the allowed execution time because when i have an array of for example size 8x3, my algorithm needs 8! comparisons when it definitely could be optimized to a much smaller number.

这也是我在CS历经3年后在这里的第一篇文章，我像一百万次一样狂暴地放弃了这个问题，所以我承认失败了，并决定向您寻求帮助:

Also its my first post here after 3 years of CS, i rage-quit this problem like a million times so i admitted defeat and decided ask you stackoverflow gods for help :>

推荐答案

我认为您可以分两个步骤进行操作，首先，从一组n中生成k个元素的组合，然后打印每个组合的排列.我测试了这段代码并正常工作:

I think you can do it in two steps, first, generate combination of k elements out of a set of n, then print permutation of each combination. I tested this code and works fine:

#include <iostream>
using namespace std;

void printArr(int a[], int n, bool newline = true) {
    for (int i=0; i<n; i++) {
        if (i > 0) cout << ",";
        cout << a[i];
    }
    if (newline) cout << endl;
}

// Generating permutation using Heap Algorithm
void heapPermutation(int a[], int n, int size) {
    // if size becomes 1 then prints the obtained permutation
    if (size == 1) {
        printArr(a, n);
        return;
    }

    for (int i=0; i<size; i++) {
        heapPermutation(a, n, size-1);
        // if size is odd, swap first and last element, otherwise swap ith and last element
        swap(a[size%2 == 1 ? 0 : i], a[size-1]);
    }
}

// Generating permutation using Heap Algorithm
void heapKPermutation(int a[], int n, int k, int size) {
    // if size becomes 1 then prints the obtained permutation
    if (size == n - k + 1) {
        printArr(a + n - k, k);
        return;
    }

    for (int i=0; i<size; i++) {
        heapKPermutation(a, n, k, size-1);
        // if size is odd, swap first and last element, otherwise swap ith and last element
        swap(a[size%2 == 1 ? 0 : i], a[size-1]);
    }
}

void doKCombination(int a[], int n, int p[], int k, int size, int start) {
    int picked[size + 1];
    for (int i = 0; i < size; ++i) picked[i] = p[i];
    if (size == k) {
        // We got a valid combination, use the heap permutation algorithm to generate all permutations out of it.
        heapPermutation(p, k, k);
    } else {
        if (start < n) {
            doKCombination(a, n, picked, k, size, start + 1);
            picked[size] = a[start];
            doKCombination(a, n, picked, k, size + 1, start + 1);
        }
    }
}

// Generate combination of k elements out of a set of n
void kCombination(int a[], int n, int k) {
    doKCombination(a, n, nullptr, k, 0, 0);
}

int main()
{
    int a[] = {1, 2, 3, 4, 5};
    cout << "n=1, k=1, a=";
    printArr(a, 1);
    kCombination(a, 1, 1);

    cout << "n=2, k=1, a=";
    printArr(a, 2);
    kCombination(a, 2, 1);

    cout << "n=3, k=2, a=";
    printArr(a, 3);
    kCombination(a, 3, 2);

    cout << "n=5, k=2, a=";
    printArr(a, 5);
    kCombination(a, 5, 2);
    return 0;
}

结果是:

n=1, k=1, a=1
1
n=2, k=1, a=1,2
2
1
n=3, k=2, a=1,2,3
2,3
3,2
1,3
3,1
1,2
2,1
n=5, k=2, a=1,2,3,4,5
4,5
5,4
3,5
5,3
3,4
4,3
2,5
5,2
2,4
4,2
2,3
3,2
1,5
5,1
1,4
4,1
1,3
3,1
1,2
2,1

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