Question

我想知道是否有一种算法可以合并满足谓词的集合中的元素类似于以下代码：

struct Element
{
  int size;
  constexpr static int maxSize = 150;
  enum Type { Mergeable, notMergeable } type;
};

auto predicate = [](const Element& el1, const Element& el2)
{
  //Done by the algorithm

  if(&el1 == &el2) // An element should not merge with itself obviously
    return false;

  if(el1.size == 0 || el2.size == 0) //element is marked for deletion
    return false; 

  //User predicate:
  bool isBelowMaxSize = el1.size + el2.size < Element::maxSize;
  bool isMergeable = el1.type  == Element::Type::Mergeable  && el2.type  == Element::Type::Mergeable;

  return isBelowMaxSize && isMergeable;
}

//User merge function
auto merge = [](Element& el1, Element& el2)
{
  el1 += el2;

  //Done by the algorithm
  el2.size = 0; //Marks for deletion
}

int main()
  std::vector<Element> els;
  //Let's assume els contains elements
  for(auto& el1 : els)
      for(auto& el2 : els)
          if(predicate(el1, el2))
            merge(el1, el2)

  //Merged elements are now removed
}

我以为我可以对范围做同样的事情：

    namespace rv = ranges::views;
    auto result = rv::cartesian_product(els, els) | rv::filter(predicate) | rv::for_each(merge);

但是我担心它无法正常工作，因为它可能会尝试合并已经合并的元素。

那么，有没有一种干净的方法？

Answer 1

您可以通过首先将所有Element::Type::Mergeable和size < Element::maxSize的{{1}}项过滤到一个单独的容器中（这是 O（n）），然后按大小排序（ O（n log n））。

有了分类的候选容器，您可以轻松地从两端进行遍历，直到迭代器在中间相遇为止。组合最大和最小元素将在线性时间内给出所有允许的合并。

Answer 2

感谢@ Useless，@ Ted Lyngmo和@Caleth的帮助，这是我的答案

//Expects a range sorted in descending order
template<class It, class Predicate, class Sum>
[[nodiscard]] static It merge_if(It first, It last, Predicate predicate, Sum sum)
{
    assert(first != last);
    last--;

    while (first != last)
    {
        while (predicate(*first, *last))
        {
            *first = sum(*first, *last);
            last--;
        }
        first++;
    }
    first++;
    return first; //One past the end
}

template<class It, class Predicate>
[[nodiscard]] static It merge_if(It first, It last, Predicate predicate)
{
    using Type = std::remove_cv_t<std::remove_reference_t<decltype(*first)>>;
    auto plus = [](const Type &first, const Type &second) { return first + second; };
    return merge_if(first, last, predicate, plus);
}

my GitLab上有一些测试用例

合并集合中满足谓词的元素

2 个答案: