在编译时确定最不常见的祖先

Question

关于最不常见的祖先算法有很多问题,但是这个问题是不同的,因为我试图在编译时确定LCA,而我的树既不是二进制也不是搜索树,即使我的简化版本可能看起来像一.

假设你有一堆包含成员typedef的结构parent,这是另一个类似的结构:

struct G
{
    typedef G parent;    // 'root' node has itself as parent
};

struct F
{
    typedef G parent;
};

struct E
{
    typedef G parent;
};

struct D
{
    typedef F parent;
};

struct C
{
     typedef F parent;
};

struct B
{
    typedef E parent;
};

struct A
{
     typedef E parent;
};

它们共同组成了一棵树

A    B    C    D
 \  /      \  /
  E         F
   \       /
    \     /
     \   /
       G

注意:结构之间没有继承关系.

我想要做的是创建一个类型特征least_common_ancestor,使得:

least_common_ancestor<A, B>::type;    // E
least_common_ancestor<C, D>::type;    // F
least_common_ancestor<A, E>::type;    // E
least_common_ancestor<A, F>::type;    // G

最好的方法是什么？

我并不关心算法的复杂性,特别是因为树的深度很小,而是我正在寻找能够得到正确答案的最简单的元程序.

编辑:我需要能够使用msvc2013和其他编译器构建解决方案,所以没有constexpr首选答案.

Answer 1

这可能会有所改进,但您可以先计算类型的深度,然后使用此信息上升到一个分支或另一个分支:

template <typename U, typename = typename U::parent>
struct depth {
    static const int value = depth<typename U::parent>::value + 1;
};

template <typename U>
struct depth<U, U> {
    static const int value = 0;
};

以上将基本计算树中类型的深度.

然后你可以使用std::enable_if:

template <typename U, typename V, typename Enabler = void>
struct least_common_ancestor;

template <typename U>
struct least_common_ancestor<U, U> {
    using type = U;
};

template <typename U, typename V>
struct least_common_ancestor<U, V,
                             typename std::enable_if<(depth<U>::value < depth<V>::value)>::type> {
    using type = typename least_common_ancestor<U, typename V::parent>::type;
};

template <typename U, typename V>
struct least_common_ancestor<U, V,
                             typename std::enable_if<(depth<V>::value < depth<U>::value)>::type> {
    using type = typename least_common_ancestor<V, typename U::parent>::type;
};

template <typename U, typename V>
struct least_common_ancestor<U, V,
                             typename std::enable_if<!std::is_same<U, V>::value && (depth<V>::value == depth<U>::value)>::type> {
    using type = typename least_common_ancestor<typename V::parent, typename U::parent>::type;
};

输出:

int main(int, char *[]) {

    std::cout << std::is_same<least_common_ancestor<A, B>::type, E>::value << std::endl;
    std::cout << std::is_same<least_common_ancestor<C, D>::type, F>::value << std::endl;
    std::cout << std::is_same<least_common_ancestor<A, E>::type, E>::value << std::endl;
    std::cout << std::is_same<least_common_ancestor<A, F>::type, G>::value << std::endl;
    std::cout << std::is_same<least_common_ancestor<A, A>::type, A>::value << std::endl;

    return 0;
}

得到:

1 1 1 1 1

这可能会有所改善,但可以作为起点.

Answer 2

template <typename...> struct typelist {};

template <typename T, typename... Ts>
struct path : path<typename T::parent, T, Ts...> {};

template <typename T, typename... Ts>
struct path<T, T, Ts...> { using type = typelist<T, Ts...>; };

template <typename T, typename U>
struct least;

template <typename T, typename... Vs, typename... Us>
struct least<typelist<T, Vs...>, typelist<T, Us...>> { using type = T; };

template <typename T, typename W, typename... Vs, typename... Us>
struct least<typelist<T, W, Vs...>, typelist<T, W, Us...>>
    : least<typelist<W, Vs...>, typelist<W, Us...>> {};

template <typename V, typename U>
using least_common_ancestor = least<typename path<V>::type, typename path<U>::type>;

DEMO

1. 自下而上:path::type通过在每个级别(path<?, T, Ts...>)前面添加一个类型列表,从两个节点到根节点路径(),直到parent等于当前处理的节点(<T, T, ?...>).向上移动是通过替换T来执行的T::parent.

2. 自上而下:同时下降两个类型列表(least),直到相应位置不匹配(Vs..., Us...); 如果是这样,最后一个公共节点是共同的祖先(T); 否则(<T, W, ?...>, <T, W, ?...>),删除匹配的节点(T)并向下一级(现在W是最后一个已知的公共节点).

Answer 3

这可能不是最具算法效率的方法,但它的功能.

首先,我们将从每种类型的祖先中创建列表.因此,对于A这将是,<A,E,G>并将G为此<G>:

template <class X>
using parent_t = typename X::parent;

template <class... > struct typelist {}; 
template <class T> struct tag_t { using type = T; };

template <class, class> struct concat;
template <class X, class Y> using concat_t = typename concat<X, Y>::type;

template <class... Xs, class... Ys> 
struct concat<typelist<Xs...>, typelist<Ys...>>
: tag_t<typelist<Xs..., Ys...>>
{ };

template <class X, class = parent_t<X>>
struct ancestors
    : tag_t<concat_t<typelist<X>, typename ancestors<parent_t<X>>::type>>
{ };

template <class X>
struct ancestors<X, X>
    : tag_t<typelist<X>>
{ };

template <class X>
using ancestors_t = typename ancestors<X>::type;

那么两个节点的最不常见的祖先就是一个节点的祖先中的第一个节点,它包含在另一个节点的祖先中:

template <class X, class TL> struct contains;
template <class X, class TL> using contains_t = typename contains<X, TL>::type;

template <class X, class... Xs>
struct contains<X, typelist<X, Xs...>> : std::true_type { };

template <class X, class Y, class... Xs>
struct contains<X, typelist<Y, Xs...>> : contains_t<X, typelist<Xs...>> { };

template <class X>
struct contains<X, typelist<>> : std::false_type { };

template <class X, class Y>
struct lca_impl;

template <class X, class Y>
struct lca : lca_impl<ancestors_t<X>, ancestors_t<Y>> { };

template <class X, class... Xs, class TL>
struct lca_impl<typelist<X, Xs...>, TL>
    : tag_t<
        typename std::conditional_t<contains_t<X, TL>::value,
            tag_t<X>,
            lca_impl<typelist<Xs...>, TL>
            >::type
        >
{ };


template <class X, class Y>
using lca_t = typename lca<X, Y>::type;

有你期望的行为:

static_assert(std::is_same<lca_t<A, E>, E>{}, "!");
static_assert(std::is_same<lca_t<A, D>, G>{}, "!");