CGAL：帮助从Delaunay Triangulation获取三角形坐标 [英] CGAL: Help getting triangles coordinates from Delaunay Triangulation

问题描述

我是CGAL的新手，我敢肯定我的问题很简单。

I'm new with CGAL, i'm sure my question is very simple.

我正在尝试使用CGAL进行一些Delaunay三角剖分。我有一个球体上具有N个3D点的网格，我想使用这些点作为三角形的顶点对球体进行三角剖分。我只需要得到这样一个三角形的顶点列表：

I am trying to use CGAL to do some Delaunay triangulation. I have a grid with N 3D points over a sphere and I want to triangulate the sphere using those points as the vertex of the triangles. I just need to get a list of the vertex of the resulting triangles like that:

id_triangle1 vertex_1 vertex_2 vertex_3
id_triangle2 vertex_1 vertex_2 vertex_3
.. .....

id_triangle1 vertex_1 vertex_2 vertex_3 id_triangle2 vertex_1 vertex_2 vertex_3 .......

我这样做是为了执行三角剖分：

I have done that to perform the triangulation:

    std::vector<Point> P; 
    for(i=0;i<NSPOINTS;i++) 
            P.push_back(Point(GRID[i].x,GRID[i].y,GRID[i].z)); 

    // building Delaunay triangulation. 
    Delaunay dt(P.begin(), P.end()); 

我的问题是我不知道如何得到结果三角剖分。我想出了如何获取face_iterator，但我不知道该怎么做：

The problem I have is that I have no idea how to get the resulting triangulation. I figured out how to get the face_iterator, but I don't know what to do from there:

    Delaunay::Finite_faces_iterator it; 
    for (it = dt.finite_faces_begin(); it != dt.finite_faces_end(); it++){ 
            std::cout << dt.triangle(it) << std::endl; 
    } 

我不确定在三角形上进行迭代是否正确，并且如果它是...
a三角形=面??¿，我的意思是，每个迭代器位置只有一个三角形？
如何正确获得每个三角形的x，y和z ??

I'm not sure if that is correct to iterate over the triangles, and if it is... a triangle = face ??¿ , i mean , each iterator position have only a triangle¿? How can I get correctly the x,y and z of each triangle¿?¿

推荐答案

如所记载的此处中，构面是一对（Cell_handle，int）。整数表示与构面相反的像元的索引。因此可以像这样完成对构面点的访问： it-> first-> vertex（（it-> second + k）％4）-> point（）且k = 1-> 3。

As documented here, a Facet is a pair (Cell_handle,int). The integer indicate the index of the cell opposite to the facet. So getting access to the points of the facet can be done like this: it->first->vertex( (it->second+k)%4 )->point() with k=1->3.

请注意，如果您对球的三角剖分（即表面三角剖分）感兴趣，则需要仅考虑入射到无限单元的多个方面。
另外，使用凸包可以解决此问题，请参见此示例。

Note that if you are interested in a triangulation of a sphere (i.e. a surface triangulation), you need to consider only facets incident to an infinite cell. Also, using the convex-hull solves this problem, see this example.

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