Line segment intersection 2d线段交叉计算 给定平面内的一组线段,求出这组线段相交的那些点。线段的数据结构是首尾的两个点。 显然,n个线段的交叉的点点个数最少0个,最多 C2n = O(n2) 个。 最简单的办法就是强行枚举所有可能,是个O (n2) 的算法,但这种方法在碰撞点很少的情况下不是很划算。所以其实是需...
Correcting self-intersecting polygons using minimal memory A simple and efficient algorithm and thoughts on line-segment intersection algorithmsComputer Science - Computational GeometryWhile well-known methods to list the intersections of either a list of segments or a complex polygon aim at achieving ...
1.Linesegmentintersection:givennlinesegments,reporttheirintersectionsefficiently.Worstcase:k=n(n–1)/2=O(n2)intersections.Optimalalgorithm:O(nlogn+k)timeandO(n)space.4 Familyofintersectionproblems 1.PolygonintersectionIntersectionoftwosimple 89 polygonsisnotasimple polygon.5 Boolean...
Line Segment Intersection 普林斯顿大学算法课http://algs4.cs.princeton.edu/home/ 反正不管怎么写也没人看这里嘛。。。
Algorithms Non-Lecture O: Line Segment Intersection Fortunately, we don’t need (or want!) to use a full-fledged convex hull algorithm just to test two segments; there’s a much simpler test. Two segments ab and cd intersect if and only if ...
Line segment intersection. In Computational Geometry: Algorithms and Ap- plications, chapter 2, pages 18-43. Springer-Verlag, 2000.De Berg, M.; van Kreveld, M.; Overmars, M.; Schwarzkopf, O. Line segment intersection. In Computational Geometry: Algorithms and Applications; Springer-Verlag: ...
ES-segment管理 首先明确几个概念: 1.segment是在磁盘上的 2.ES删除数据导致磁盘容量上升原因:ES才用的标记删除,首先会将要合并的数据拷贝出来,重新写入到新的segment中,然后删除旧的数据,所以会导致消耗额外的磁盘和IO 3.ReFresh:从文件系统缓存中把数据写入到磁盘segment,并打开segment,使得新数据可以搜索的过程...
(LineRow1,LineColumn1) and (LineRow2,LineColumn2). The intersection point, if it exists, is returned in (Row,Column). If the segment and the line have a part in common,IsOverlappingreturns the value 1, otherwise 0 is returned. In this case the endpoints of the mutual segment are ...
We present an optimal cache-oblivious algorithm for finding all intersections between a set of non-intersecting red segments and a set of non-intersecting blue segments in the plane. Our algorithm uses/memory transfers, whereNis the total number of segments,MandBare the memory and block transfer ...
Well, first parameterize the line segment by a parameter T. x = x1+(x2-x1)*T y = y1+(y2-y1)*T The line segment goes from T = 0 --> 1. Assume the line segment intersects the line y=mx+c as some time t. Then [y1+(y2-y1)*t] = m*[x1+(x2-x1)*t] + c ...