可能重复：   How我可以创建载体载体的笛卡尔积？

我在搞清楚如何生成元素的所有组合中的2D矢量一些逻辑问题。在这里，我创建了一个二维矢量。既不尺寸的大小可以假设

I'm having some logical issues figuring out how to generate all combinations of elements in a 2d vector. Here, I create a 2D vector. Neither dimension's size can be assumed.

#include <iostream>
#include <vector>

using namespace std;

int main() {
  srand(time(NULL));
  vector< vector<int> > array;

  // This creates the following:
  // array[0]: {0, 1, 2} 
  // array[1]: {3, 4, 5, 9} 
  // array[2]: {6, 7, 8} 
  for(int i=0; i<3; i++) { 
    vector<int> tmp;
    tmp.push_back((i*3)+0); tmp.push_back((i*3)+1); tmp.push_back((i*3)+2);
    if(i==1)
      tmp.push_back((i*3)+6);
    array.push_back(tmp);
  }
}

创建载体后，我想输出所有可能的组合如下：

After creating the vector, I'd like to output all possible combinations as follows:

  comb[0] = {0, 3, 6}
  comb[1] = {0, 3, 7}
  comb[2] = {0, 3, 8}
  comb[3] = {0, 4, 6}
  comb[4] = {0, 4, 7}
  comb[x] = {...}

然而，我无法如何概念化环路结构做这个正常，其中该尺寸'阵列'和每个子阵列中的元素是未知的/动态的。

However, I am having trouble how to conceptualize the loop structure to do this properly, where the size 'array' and the elements in each sub-array is unknown/dynamic.

编辑1：不能假定有3个阵列。有array.size其中（））

EDIT 1: Can't assume there are 3 arrays. There are array.size() of them ;)

推荐答案

未知大小的最简单的方法就是递归。

The easiest way for unknown sizes is recursion.

void combinations(vector<vector<int> > array, int i, vector<int> accum)
{
    if (i == array.size()) // done, no more rows
    {
        comb.push_back(accum); // assuming comb is global
    }
    else
    {
        vector<int> row = array[i];
        for(int j = 0; j < row.size(); ++j)
        {
            vector<int> tmp(accum);
            tmp.push_back(row[j]);
            combinations(array,i+1,tmp);
        }
    }
}

起初与调用I = 0 和一个空的 ACCUM 。