※ 引述《makiyolove (暴力熊)》之铭言： : 各位版友好： : 在一起)。现在我想要在这个不规则图形上找重心，但是找 google 了一下看到这个.. http://www.exaflop.org/docs/cgafaq/cga2.html 我还看完了，顺便帮你整理出重点： 一整个多边形的重心好似很难算， 但是一个三角形的重心很好算， 两个三角形的重心也很好算，三个三角型的重心也很好算， 四个也很好算…五个也…六个也…七个也…八格也路… 原来如此！我们只要把多边形三角化以後， 求所有三角形的重心的重心就好了！ 报告完毕。 我人真好 QQ -- Subject 2.02: How can the centroid of a polygon be computed? The centroid (a.k.a. the center of mass, or center of gravity) of a polygon can be computed as the weighted sum of the centroids of a partition of the polygon into triangles. The centroid of a triangle is simply the average of its three vertices, i.e., it has coordinates (x1 + x2 + x3)/3 and (y1 + y2 + y3)/3. This suggests first triangulating the polygon, then forming a sum of the centroids of each triangle, weighted by the area of each triangle, the whole sum normalized by the total polygon area. This indeed works, but there is a simpler method: the triangulation need not be a partition, but rather can use positively and negatively oriented triangles (with positive and negative areas), as is used when computing the area of a polygon. This leads to a very simple algorithm for computing the centroid, based on a sum of triangle centroids weighted with their signed area. The triangles can be taken to be those formed by one fixed vertex v0 of the polygon, and the two endpoints of consecutive edges of the polygon: (v1,v2), (v2,v3), etc. The area of a triangle with vertices a, b, c is half of this expression: (b[X] - a[X]) * (c[Y] - a[Y]) - (c[X] - a[X]) * (b[Y] - a[Y]); Code available at ftp://grendel.csc.smith.edu/pub/code/centroid.c (3K). Reference: [Gems IV] pp.3-6; also includes code. -- To iterate is human, to recurse, divine. 递回只应天上有, 凡人该当用回圈. 　 L. Peter Deutsch --

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