使用提升几何缓冲区缩放多边形时的冗余折点

Question

我正在使用提升几何来管理一些多边形，我需要按给定的数量扩展和缩小它们。我正在使用 boost：：geometry：：buffer 来完成此操作，我想知道是否有更好的选择。我担心的是，如果我扩展一个矩形，我最终会得到一个有 12 个顶点的多边形(其中 8 个无关紧要(。下面是一些示例代码：

#include <iostream>
#include <boost/geometry.hpp>
#include <boost/geometry/geometries/geometries.hpp>
#include <boost/geometry/geometries/point_xy.hpp>
using point_t = boost::geometry::model::d2::point_xy<double>;
using polygon_t = boost::geometry::model::polygon<point_t>;
using mpolygon_t = boost::geometry::model::multi_polygon<polygon_t>;
mpolygon_t expand_polygon(const polygon_t& polygon,
const double distance) {
mpolygon_t scaled_polygon;
// DistanceStrategy is symmetric; want straight lines (no curves)
boost::geometry::strategy::buffer::distance_symmetric<double>
distance_strategy{distance};
// SideStrategy is straight; grow equally on all sides
boost::geometry::strategy::buffer::side_straight side_strategy;
// JoinStrategy is miter; 1.0 as limit; Sharp (not rounded) joins
boost::geometry::strategy::buffer::join_miter join_strategy;
// EndStrategy is flat; flat (not rounded) ends
boost::geometry::strategy::buffer::end_flat end_strategy;
// PointStrategy is square; squares, not circles, if poly is just a point
boost::geometry::strategy::buffer::point_square point_strategy;
boost::geometry::buffer(polygon, scaled_polygon,
distance_strategy,
side_strategy,
join_strategy,
end_strategy,
point_strategy);
// return scaled polygon offset by supplied distance
return scaled_polygon;
}
int main() {
using boost::geometry::get;
polygon_t rect;
boost::geometry::read_wkt("POLYGON((5 5,5 8,8 8,8 5, 5 5))", rect);
mpolygon_t expanded_mpoly = expand_polygon(rect, 1.0);
auto list_coordinates = [](const point_t& pt) {
std::cout.precision(std::numeric_limits<double>::max_digits10);
std::cout << "vertex(" << get<0>(pt)
<< ", " << get<1>(pt) << ")" << std::endl;
};
boost::geometry::for_each_point(expanded_mpoly, list_coordinates);
return 0;
}

这是输出：

vertex(4, 5)
vertex(4, 8)
vertex(4, 9)
vertex(5, 9)
vertex(8, 9)
vertex(9, 9)
vertex(9, 8)
vertex(9, 5)
vertex(9, 4)
vertex(8, 4)
vertex(5, 4)
vertex(4, 4)
vertex(4, 5)

我需要提供策略参数吗？我应该考虑其他一些提升的领地？

Answer 1

Boost 包含simplify算法(它使用 Ramer-Douglas-Peucker 算法(。例如在你的多边形中有3个点p1,p2,p3，我们计算点p2和线|p1p3|之间的距离，如果这个距离小于给定的epsilon值，p2点从多边形中删除。因此，如果您使用epsilon值非常小的函数(以移除共线点(，则可以从多边形的所有边移除所有冗余点。试试这个

mpolygon_t out; // to hold new geometry
boost::geometry::simplify(expanded_mpoly,out,0.00001); // very small epsilon value
auto list_coordinates = [](const point_t& pt) {
std::cout.precision(std::numeric_limits<double>::max_digits10);
std::cout << "vertex(" << get<0>(pt)
<< ", " << get<1>(pt) << ")" << std::endl;
};
boost::geometry::for_each_point(out, list_coordinates); // print out

作为您看到的输出

vertex(9, 9)
vertex(9, 4)
vertex(4, 4)
vertex(4, 9)
vertex(9, 9)