从Projection_traits_xy_3的约束Delaunay三角剖分中检索z [英] Retrieve z from a constrained delaunay triangulation of Projection_traits_xy_3

问题描述

给出 x，y ，如何检索使用 projection_traits_xy_3 从2.5D构建的2D约束Delaunay三角剖分的 z 坐标?/p>

Given x,y, how can I retrieve the z coordinate of a 2D constrained delaunay triangulation built from 2.5D with projection_traits_xy_3?

typedef CGAL::Exact_predicates_inexact_constructions_kernel K;
typedef CGAL::Projection_traits_xy_3<K>  Gt;

typedef CGAL::Triangulation_vertex_base_2<Gt> Vb;
typedef CGAL::Delaunay_mesh_face_base_2<Gt> Fb;
typedef CGAL::Triangulation_data_structure_2<Vb, Fb> Tds;
typedef CGAL::Constrained_Delaunay_triangulation_2<Gt, Tds> CDT;

我的猜测是我必须取回脸，但是下一步将是什么?

My guess is that I have to retrieve the face, but what would be the next step?

CDT::Point query(10,10,?);   
CDT::Face_handle face_handle = cdt.locate(query);

推荐答案

正如@Andreas指出的那样，三角剖分即使使用2D投影也可以存储3D点.因此，我们可以检索 Point_3 .

As @Andreas pointed out, the triangulation stores 3d points even if it uses its 2d projection. Therefore, we can retrieve the Point_3.

给出查询点(x，y)=(100,100):

//z unknown
Point query1(100, 100, 0);

CDT::Face_handle face_handle = cdt.locate(query1);

K::Point_3 p = face_handle->vertex(0)->point();
K::Point_3 q = face_handle->vertex(1)->point();
K::Point_3 r = face_handle->vertex(2)->point();

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