Question

假设我有一个非简单的多边形， CGAL如何帮助我将其划分为一组简单的多边形？

例如，给出由一系列2D点表示的多边形：

(1, 1) (1, -1) (-1, 1) (-1, -1)

我希望获得两个多边形;

(1, 1) (1, -1) (0, 0)

和

(0, 0) (-1, 1) (-1, -1)

CGAL是否可行？

Answer 1

您需要的两个多边形不构成原始船体。如果您只想使用（0,0）作为顶点之一将原始集合分成三角形，则可以执行此操作：

#include <CGAL/Exact_predicates_inexact_constructions_kernel.h>
#include <CGAL/Constrained_Delaunay_triangulation_2.h>
#include <CGAL/Delaunay_mesh_vertex_base_2.h>
#include <CGAL/Delaunay_mesh_face_base_2.h>
#include <vector>    

typedef CGAL::Exact_predicates_inexact_constructions_kernel     K;
typedef K::Point_2                                              Point_2;
typedef CGAL::Delaunay_mesh_vertex_base_2<K>                    Vb;
typedef CGAL::Delaunay_mesh_face_base_2<K>                      Fb;
typedef CGAL::Triangulation_data_structure_2<Vb, Fb>            Tds;
typedef CGAL::Constrained_Delaunay_triangulation_2<K, Tds>      CDT;
typedef CDT::Vertex_handle                                      Vertex_handle;
typedef CDT::Face_iterator                                      Face_iterator;    

int main(int argc, char* argv[])
{
    // Create a vector of the points
    //
    std::vector<Point_2> points2D ;
    points2D.push_back(Point_2(  1,  1));
    points2D.push_back(Point_2(  1, -1));
    points2D.push_back(Point_2( -1,  1));
    points2D.push_back(Point_2( -1, -1));
    points2D.push_back(Point_2( 0, 0));

    size_t numTestPoints = points2D.size();

    // Create a constrained delaunay triangulation and add the points
    //
    CDT cdt;
    std::vector<Vertex_handle> vhs;
    for (unsigned int i=0; i<numTestPoints; ++i){
        vhs.push_back(cdt.insert(points2D[i]));
    }

    int i=0;
    for (Face_iterator fit = cdt.faces_begin()  ; fit != cdt.faces_end(); ++fit) {
        printf("Face %d is (%f,%f) -- (%f,%f) -- (%f,%f) \n",i++,
               fit->vertex(0)->point().x(),fit->vertex(0)->point().y(),
               fit->vertex(1)->point().x(),fit->vertex(1)->point().y(),
               fit->vertex(2)->point().x(),fit->vertex(2)->point().y() );

    }

    return 0 ;
}

哪个应该给你这样的输出：

Face 0 is (0.000000,0.000000) -- (1.000000,-1.000000) -- (1.000000,1.000000) 
Face 1 is (0.000000,0.000000) -- (1.000000,1.000000) -- (-1.000000,1.000000) 
Face 2 is (-1.000000,-1.000000) -- (0.000000,0.000000) -- (-1.000000,1.000000) 
Face 3 is (-1.000000,-1.000000) -- (1.000000,-1.000000) -- (0.000000,0.000000)

使用CGAL分割非简单多边形

1 个答案: