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这是一个相当补救的问题。我一直在查看 JTS DelaunayTriangulationBuilder 的文档，我不知道如何做看起来应该很简单的事情。我希望获取一系列点，对它们进行三角剖分，然后在该网格中插入随机点的 Z 值。粗略阅读如何做到这一点并不明显。有任何想法吗？

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2 回答 2

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加载三角剖分对象后，调用getSubdivision()它来获取三角剖分。它使用四边数据结构，稍后您将需要它。（如果您知道什么是半边或翼边表示，则更容易理解。）结果QuadEdgeSubdivision有一个方法locate，给定一个坐标，返回封闭三角形的一条边（作为四边形）。用 获取它的源顶点，用 获取它orig()的目标顶点dest()。获取另一条边，oNext()其目标顶点是第三个顶点（dPrev().origin() 也是同一个顶点）。现在您有了三个顶点，将您的测试点表示为 aVertex和 call interpolateZValue。

例如：

public static double 
interpolateZ(DelaunayTriangulationBuilder triangulation,
             Coordinate coordinate) {
    QuadEdgeSubdivision quadEdgeSubdivision = triangulation.getSubdivision();
    QuadEdge edge = quadEdgeSubdivision.locate(coordinate);
    return new Vertex(coordinate.x, coordinate.y)
            .interpolateZValue(edge.orig(), edge.dest(), edge.oNext().dest());
}

不过，你是对的。通过阅读他们的 API 如何做到这一点并不明显。

于 2012-11-20T02:24:08.577 回答
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I'm not familiar with JTS DelauneyTriangulationBuilder, but it sounds like you have a collection of points (x,y,z), and you are submitting the 2D (x,y) pairs to the triangulator. This gives you a planar triangulation of the (x,y) points, but also a mesh who's vertices are the original (x,y,z) points.

Once you have a triangulation, you wish to find the point (p,q,r) on the mesh that corresponds to the planar point (p,q). To do so, find the triangle T of the Delauney triangulation that contains (p,q). Find the barycentric coordinates of (p,q) relative to T, and use these to compute a weighted average r of the z values corresponding to the vertices of T. That weighted average is the Z value you're looking for. In other words, (p,q,r) is on the mesh.

于 2012-11-05T19:59:49.393 回答