line sweep algorithm java

Left edge is represented by lower-left point and right edge by upper-right point.
This naive implementation is really easy to implement : public static Point naiveClosestPair(Point points) double min X_value; Point closestPair new Point2; for (Point p1 : points) for (Point p2 : points) if (p1!
Writing code in comment?This step takes O(log N).Therefore, this approach takes O(n2).With a Self-Balancing BST, we can do all of the above operations in O(Logn) time.In this article I'll explain an efficient algorithm using plane sweep, compare it to the naive implementation and discuss its complexity.A rectangle is represented by two points, one lower-left point and one upper-right point.So, the overall complexity of this loop is O(N * log N).And add its line to active line segments (line segments for which photo contests free left end point is seen, but right end point is not seen yet). All points from left to right are processed one by one.
Fetching contributors, cannot retrieve contributors at this time * Copyright (c) 2010, Bart Kiers * * Permission is hereby granted, free of charge, to any person * obtaining a copy of this software and associated documentation * files (the "Software to deal in the Software.At any instance, the set contains only the rectangles which intersect the sweep line (rectangles whose left edges are visited but right edges are not).Type0) /If it is a bottom edge of rectangle if (cnt0) begin_y rectsevents_d0.y; / Block starts cnt; rebates for tankless water heaters 2014 /incrementing number of overlapping rectangles else /If it is a top edge -cnt; /the rectangle is no more overlapping, so remove it if (cnt0) /Block ends int [email protected] segments the line segments.We start our algorithm by sorting the events by x coordinates.Here we are given n line segments and we need to find out if any two line segments intersect or not.Do explore this technique's application in some other problems.2 is inserted into the Tree.