我使用boost几何模型创建了一个线段。我想知道是否有一种方法可以将分段划分为N个相等的线段。
请不要将此问题视为this的重复版本。 (修改) 因为过时了,我已经尝试过为这个问题提供的答案。
答案 0 :(得分:1)
想到最简单的事情:
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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)
对于大N来说可能更好的扩展:
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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)
答案 1 :(得分:0)
如果需要将AB段划分为N个相等的块,则可以使用线的参数方程计算第i个段(以及上一个段的末尾)的起点坐标。
伪代码:
for i = 0.. N-1
t = i / N
Start[i].X = A.X * (1 - t) + B.X * t
Start[i].Y = A.Y * (1 - t) + B.Y * t