c++ - 使用boost geometry将boost段分成N个相等的段

Question

我使用 boost 几何模型创建了一条线段。我想知道是否有一种方法可以在 boost 中将线段划分为 N 个相等的线段。

请不要将此问题视为 this 的重复版本.(编辑)因为已经过时了，我已经尝试过为这个问题提供的答案。

Answer 1

想到的最简单的事情:

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#include <boost/geometry/geometry.hpp>
#include <boost/geometry/io/io.hpp>
#include <boost/geometry/arithmetic/arithmetic.hpp>
#include <boost/geometry/geometries/segment.hpp>
#include <boost/geometry/geometries/point_xy.hpp>
#include <boost/geometry/geometries/linestring.hpp>

#include <iostream>

template <typename Segment, typename Point = typename boost::geometry::point_type<Segment>::type>
auto split_into(unsigned N, Segment const& s) {
    using namespace boost::geometry;
    assert(N>1);

    auto step = s.second;
    subtract_point(step, s.first);
    divide_value(step, N-1);

    std::vector<Point> segments;
    std::generate_n(back_inserter(segments), N, [accu=s.first, step] () mutable { Point p = accu; return add_point(accu, step), p; });

    return model::linestring<Point> {segments.begin(), segments.end()};
}

using namespace boost::geometry;
using Point = model::d2::point_xy<double>;

int main() {
    model::segment<Point> const line { {1,3}, {11,33} };

    std::cout << wkt(line) << "\n";
    std::cout << wkt(split_into(5, line)) << "\n";
}

打印

LINESTRING(1 3,11 33)
LINESTRING(1 3,3.5 10.5,6 18,8.5 25.5,11 33)

花式范围版本

对于大 N 可能更好地扩展:

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#include <boost/range/adaptors.hpp>
#include <boost/range/irange.hpp>

template <typename Segment, typename Point = typename boost::geometry::point_type<Segment>::type>
auto split_into(unsigned N, Segment const& s) {
    using namespace boost::geometry;
    using namespace boost::adaptors;
    assert(N>1);

    auto step = s.second;
    subtract_point(step, s.first);
    divide_value(step, N-1);

    auto gen = boost::irange(0u, N) | transformed([=](int i) { auto p = step; multiply_value(p, i); add_point(p, s.first); return p; });
    return model::linestring<Point> { boost::begin(gen), boost::end(gen) };
}

打印

LINESTRING(1 3,11 33)
LINESTRING(1 3,3.5 10.5,6 18,8.5 25.5,11 33)