将范围分为重叠范围

Split range into range of overlapping ranges

本文关键字：范围 重叠      更新时间：2023-10-16

i试图使用范围-V3库将值的容器切成一个范围，以使相邻的范围共享边界元素。

考虑以下内容：

using namespace ranges;
std::vector<int> v = { 1, 2, 3, 0, 4, 0, 5, 0, 6, 7, 8, 0, 0, 9 };
auto myRanges = v | /* something like adjacent split */
for_each( myRanges, []( auto&& range ){ std::cout << range << std::endl;} );

我想将范围分为重叠子兰，该区域是否满足两个条件：

  1. 该元素是否具有零值
  2. 或与一个或多个元素相邻，值为零

所需的输出：

[1,2,3]
[3,0,4,0,5,0,6]
[6,7,8]
[8,0,0,9]

我的尝试：

auto degenerate =
  []( auto&& arg ){
    return distance( arg ) < 2;  
  };
auto myRanges = v | view::split(0) | view::remove_if( degenerate );
for_each( myRanges, []( auto&& range ){ std::cout << range << std::endl;} );

输出：

[1,2,3]
[6,7,8]

我对我的可能如何

  1. "插入" 3到6的范围
  2. "附加"范围从8到9

如果我正确理解您的要求，则可以根据adjacent_find实现生成器：

namespace detail {
    template<typename IterT, typename SentT>
    struct seg_gen_fn {
        IterT it_;
        SentT end_;
        bool parity_ = true;
        ranges::iterator_range<IterT> operator ()() {
            if (it_ == end_) {
                return {it_, it_};
            }
            auto n = ranges::adjacent_find(
                it_, end_,
                [p = std::exchange(parity_, !parity_)](auto const a, auto const b) {
                    return a && !b == p;
                }
            );
            return {
                std::exchange(it_, n),
                n != end_ ? ranges::next(std::move(n)) : std::move(n)
            };
        }
    };
    template<typename RngT>
    constexpr auto seg_gen(RngT&& rng)
     -> seg_gen_fn<ranges::iterator_t<RngT>, ranges::sentinel_t<RngT>>
    { return {ranges::begin(rng), ranges::end(rng)}; }
} // namespace detail
auto const segmented_view = [](auto&& rng) {
    return ranges::view::generate(detail::seg_gen(decltype(rng)(rng)))
         | ranges::view::take_while([](auto const& seg) { return !seg.empty(); });
};
int main() {
    auto const ns = {1, 2, 3, 0, 4, 0, 5, 0, 6, 7, 8, 0, 0, 9};
    ranges::copy(segmented_view(ns), ranges::ostream_iterator<>{std::cout, "n"});
}

在线演示
并不像人们希望的那样简洁...： - [

对于一次性代码来说，这可能还不错，但是更多的工作，并且可以重复使用很多：

namespace detail {
    namespace tag = ranges::tag;
    template<
        typename RngT, typename PredT, typename IterT = ranges::iterator_t<RngT>,
        typename StateT = ranges::tagged_compressed_tuple<
            tag::begin(IterT), tag::end(ranges::sentinel_t<RngT>),
            tag::current(bool), tag::fun(ranges::semiregular_t<PredT>)
        >
    >
    struct seg_gen_fn : private StateT {
        constexpr seg_gen_fn(RngT&& rng, PredT pred)
          : StateT{ranges::begin(rng), ranges::end(rng), true, std::move(pred)}
        { }
        ranges::iterator_range<IterT> operator ()() {
            StateT& state = *this;
            auto& it = state.begin();
            if (it == state.end()) {
                return {it, it};
            }
            auto& parity = state.current();
            auto n = ranges::adjacent_find(
                it, state.end(),
                [p = std::exchange(parity, !parity), &pred = state.fun()]
                (auto const& a, auto const& b) {
                    return !pred(a) && pred(b) == p;
                }
            );
            return {
                std::exchange(it, n),
                n != state.end() ? ranges::next(std::move(n)) : std::move(n)
            };
        }
    };
    template<typename RngT, typename PredT>
    constexpr seg_gen_fn<RngT, PredT> seg_gen(RngT&& rng, PredT pred) {
        return {std::forward<RngT>(rng), std::move(pred)};
    }
} // namespace detail
auto const segmented_view = [](auto&& rng, auto pred) {
    return ranges::view::generate(detail::seg_gen(decltype(rng)(rng), std::move(pred)))
         | ranges::view::take_while([](auto const& seg) { return !seg.empty(); });
};
int main() {
    auto const ns = {1, 2, 3, 0, 4, 0, 5, 0, 6, 7, 8, 0, 0, 9};
    ranges::copy(
        segmented_view(ns, [](auto const n) { return n == 0; }),
        ranges::ostream_iterator<>{std::cout, "n"}
    );
}

在线演示

概念检查和预测作为练习。

相关文章：