c++ - 如何计算 N 个有序集的交集？

Question

下面的例子展示了如何计算两个集合的交集。 STL 是否提供允许不仅为 2 还为 N 执行此操作的工具？套？

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

int main()
{
    std::vector<int> v1 = { 1,2,9,3,4,5 };
    std::vector<int> v2 = { 9,4,2,7,4,1 };
    std::vector<int> v(v1.size() + v2.size());
    std::vector<int>::iterator it;

    std::sort(v1.begin(), v1.end());
    std::sort(v2.begin(), v2.end());
    it = std::set_intersection(v1.begin(), v1.end(), v2.begin(), v2.end(), v.begin());

    v.resize(it - v.begin());
    std::cout << "Both sets have " << (v.size()) << " elements in common:\n";
    for (it = v.begin(); it != v.end(); ++it)
    {
        std::cout << *it << ' ';
    }
    std::cout << '\n';

    return 0;
}

Answer 1

Does the STL provide tools that allow to do this not only for 2 butfor N sets?

不。但是您可以通过提供 recursive Variadic template 轻松制作一个如下。
if constexpr 部分需要 c++17支持。不过例子很多， how you could do it for prior to c++17 .此外，由于递归调用，参数必须以相反的顺序传递，以获得您尝试的行为。
( See Online Demo )

#include <vector>
#include <algorithm>  // std::set_intersection
#include <iterator>   // std::back_inserter

template<typename Container, typename... Rest>
Container NSetIntersections(
    const Container& container1, const Container& container2, Rest&&... rest) noexcept
{
    if constexpr (sizeof...(Rest) == 0)
    {
        Container result;
        std::set_intersection(container1.begin(), container1.end(),
            container2.begin(), container2.end(), std::back_inserter(result));
        return result;
    }
    else
    {
        Container result;
        std::set_intersection(container1.begin(), container1.end(),
            container2.begin(), container2.end(), std::back_inserter(result));
        return NSetIntersections(result, std::forward<Rest>(rest)...);
    }
}

int main()
{
    // sorted vectors
    std::vector<int> v1 = { 1, 2, 3, 4, 5, 6 };
    std::vector<int> v2 = { 2, 3, 4, 7, 8, 9 };
    std::vector<int> v3 = { 3, 4, 7, 200 };
    std::vector<int> v4 = { 4, 100, 200, 300 };
    std::vector<int> v5 = { 4, 100, 200 };
    // call the function like
    const auto res1 = NSetIntersections(v2, v1);              // 2 3 4
    const auto res2 = NSetIntersections(v3, v2, v1);          // 3 4
    const auto res3 = NSetIntersections(v4, v3, v2, v1);      // 4
    const auto res4 = NSetIntersections(v5, v4, v3, v2, v1);  // 4
    return 0;
}

为了将参数传递给 NSetIntersections以自然的方式运行，我建议遵循辅助函数的方式。另外，它还可以处理将单个参数(以防万一，错误!)传递给 NSetIntersections 的情况。 , 和 c++11兼容的。
( See Online Demo )

#include <vector>
#include <algorithm>  // std::set_intersection
#include <iterator>   // std::back_inserter

namespace helper { // helper NSetIntersections functions
    template<typename Container>
    Container NSetIntersections(const Container& container1) noexcept {
        return container1;
    }

    template<typename Container>
    Container NSetIntersections(const Container& container1, const Container& container2) noexcept
    {
        Container result;
        std::set_intersection(container1.begin(), container1.end(),
            container2.begin(), container2.end(), std::back_inserter(result));
        return result;
    }

    template<typename Container, typename... Rest>
    Container NSetIntersections(
        const Container& container1, const Container& container2, Rest&&... rest) noexcept
    {
        return helper::NSetIntersections(
            helper::NSetIntersections(container1, container2), std::forward<Rest>(rest)...);
    }
}

template<typename... Containers>
auto NSetIntersections(Containers&&... rest) noexcept
  -> decltype(helper::NSetIntersections(std::forward<Containers>(rest)...))
{
    return helper::NSetIntersections(std::forward<Containers>(rest)...);
}

现在你可以像这样使用 args 调用函数:

// sorted vectors
std::vector<int> v1 = { 1, 2, 3, 4, 5, 6 };
std::vector<int> v2 = { 2, 3, 4, 7, 8, 9 };
std::vector<int> v3 = { 3, 4, 7, 200 };
std::vector<int> v4 = { 4, 100, 200, 300 };
std::vector<int> v5 = { 4, 100, 200 };
// call the function like
const auto res1 = NSetIntersections(v1);                 // 1 2 3 4 5 6 
const auto res2 = NSetIntersections(v1, v2);             // 2 3 4 
const auto res3 = NSetIntersections(v1, v2, v3);         // 3 4 
const auto res4 = NSetIntersections(v1, v2, v3, v4);     // 4
const auto res5 = NSetIntersections(v1, v2, v3, v4, v5); // 4

旁注:在 quick-bench.com 中完成的基准测试显示(几乎)相同的性能(对于 5 个已排序的容器)，而我们本来可以做 N 次 std::set_intersection .

( See Online Quick-bench )