Question

我正在使用Boost几何来管理一些多边形，并且需要将它们扩展和缩小一定的数量。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，则最终会得到一个具有12个顶点的多边形（其中8个是不相关的）。这是一些示例代码：

#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)

是否需要提供策略论点？我应该看看其他一些增强的力量？

Answer 1

提升包含simplify算法（它使用Ramer–Douglas–Peucker algorithm）。例如，在您的多边形中有3个点p1,p2,p3，如果该距离小于给定的p2值，我们将计算点|p1p3|与线epsilon之间的距离， p2点已从多边形中删除。因此，如果您以很小的epsilon值使用此函数（以删除共线点），则可以从多边形的所有边上删除所有冗余点。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

您看到的输出

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

使用boost-geometry缓冲区缩放多边形时的冗余顶点

1 个答案: