C++向量的所有组合

Question

让我们说我们有一些数字0 to n,我们想要争夺那些规模,s并希望看到每一种可能的组合.

所以排列的数量恰好等于s! * n!/(s!*(n-s)!).

带n = 3和的示例s = 3:

0 1 2 | 0 1 3 | 0 2 1 | 0 2 3 | 0 3 1 | 0 3 2 | 1 0 2 | 1 0 3 | 1 3 2 
1 2 3 | 1 2 0 | 1 3 0 | 2 0 1 | 2 1 0 | 2 0 3 | 2 3 0 | 2 3 1 | 2 1 3
            3 0 1 | 3 1 0 | 3 0 2 | 3 2 0 | 3 1 2 | 3 2 1

使用boost/stl实现这一目标是否顺利？

Answer 1

LIVE DEMO

#include <algorithm>
#include <vector>
#include <iterator>
#include <iostream>

void dfs(int depth, int s, int i, std::vector<int>& c, const std::vector<int>& v)
{
    if (depth == s)
    {
        // base case: all chosen elements are in v, so we print
        //            all permutations of v
        do
        {
            std::copy(c.begin(), c.end(), std::ostream_iterator<int>(std::cout, " "));
            std::cout << "| ";
        }
        while (std::next_permutation(c.begin(), c.end()));
    }
    else
    {
        // regular case: loop and recurse, which effectively creates an
        //               n-dimensional for loop.
        //               This is the classic approach to finding the
        //               Cartesian product of N vectors or all combinations of elements.
        for (int j = i + 1; j < (int)v.size(); ++j)
        {
            c.push_back(v[j]);
            dfs(depth + 1, s, j, c, v);
            c.pop_back();
        }
    }
}

int main()
{
    std::vector<int> v{ 0, 1, 2, 3 };
    std::sort(v.begin(), v.end());
    v.erase(std::unique(v.begin(), v.end()), v.end());    
    std::vector<int> c;
    const int length = 3;
    dfs(0, length, -1, c, v);
}

输出：

0 1 2 | 0 2 1 | 1 0 2 | 1 2 0 | 2 0 1 | 2 1 0 | 0 1 3 | 0 3 1 | 1 0 3 |
1 3 0 | 3 0 1 | 3 1 0 | 0 2 3 | 0 3 2 | 2 0 3 | 2 3 0 | 3 0 2 | 3 2 0 |
1 2 3 | 1 3 2 | 2 1 3 | 2 3 1 | 3 1 2 | 3 2 1

Answer 2

以下是使用TC在评论中引用的链接的代码(http://howardhinnant.github.io/combinations.html):

#include "../combinations/combinations"
#include <iostream>
#include <vector>

int
main()
{
    std::vector<int> v{0, 1, 2, 3};
    typedef std::vector<int>::iterator Iter;
    for_each_permutation(v.begin(), v.begin()+3, v.end(),
        [](Iter f, Iter l)
        {
            for (; f != l; ++f)
                std::cout << *f << ' ';
            std::cout << "| ";
            return false;
        }
    );
    std::cout << '\n';
}

0 1 2 | 0 2 1 | 1 0 2 | 1 2 0 | 2 0 1 | 2 1 0 | 0 1 3 | 0 3 1 | 1 0 3 | 1 3 0 | 3 0 1 | 3 1 0 | 0 2 3 | 0 3 2 | 2 0 3 | 2 3 0 | 3 0 2 | 3 2 0 | 1 2 3 | 1 3 2 | 2 1 3 | 2 3 1 | 3 1 2 | 3 2 1 | 

这个库的一个显着优点std::next_permutation是被置换的元素不需要进行排序,甚至也不需要低于可比性.例如:

#include "../combinations/combinations"
#include <iostream>
#include <vector>

enum class color
{
    red,
    green,
    blue,
    cyan
};

std::ostream&
operator<< (std::ostream& os, color c)
{
    switch (c)
    {
    case color::red:
        os << "red";
        break;
    case color::green:
        os << "green";
        break;
    case color::blue:
        os << "blue";
        break;
    case color::cyan:
        os << "cyan";
        break;
    }
    return os;
}

int
main()
{
    std::vector<color> v{color::blue, color::red, color::cyan, color::green};
    typedef std::vector<color>::iterator Iter;
    for_each_permutation(v.begin(), v.begin()+3, v.end(),
        [](Iter f, Iter l)
        {
            for (; f != l; ++f)
                std::cout << *f << ' ';
            std::cout << "| ";
            return false;
        }
    );
    std::cout << '\n';
}

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