标签:nbsp output other arc cte reac stat run sys
we start from the leftmost point (or point with minimum x coordinate value) and we keep wrapping points in counterclockwise direction. The big question is, given a point p as current point, how to find the next point in output? The idea is to use orientation() here. Next point is selected as the point that beats all other points at counterclockwise orientation, i.e., next point is q if for any other point r, we have “orientation(p, r, q) = counterclockwise”. Following is the detailed algorithm.
1) Initialize p as leftmost point.
2) Do following while we don’t come back to the first (or leftmost) point.
…..a) The next point q is the point such that the triplet (p, q, r) is counterclockwise for any other point r.
…..b) next[p] = q (Store q as next of p in the output convex hull).
…..c) p = q (Set p as q for next iteration).
import java.util.*; class Point { int x, y; Point(int x, int y){ this.x=x; this.y=y; } } class GFG { // To find orientation of ordered triplet (p, q, r). // The function returns following values // 0 --> p, q and r are colinear // 1 --> Clockwise // 2 --> Counterclockwise public static int orientation(Point p, Point q, Point r) { int val = (q.y - p.y) * (r.x - q.x) - (q.x - p.x) * (r.y - q.y); if (val == 0) return 0; // collinear return (val > 0)? 1: 2; // clock or counterclock wise } // Prints convex hull of a set of n points. public static void convexHull(Point points[], int n) { // There must be at least 3 points if (n < 3) return; // Initialize Result Vector<Point> hull = new Vector<Point>(); // Find the leftmost point int l = 0; for (int i = 1; i < n; i++) if (points[i].x < points[l].x) l = i; // Start from leftmost point, keep moving // counterclockwise until reach the start point // again. This loop runs O(h) times where h is // number of points in result or output. int p = l, q; do { // Add current point to result hull.add(points[p]); // Search for a point ‘q‘ such that // orientation(p, x, q) is counterclockwise // for all points ‘x‘. The idea is to keep // track of last visited most counterclock- // wise point in q. If any point ‘i‘ is more // counterclock-wise than q, then update q. q = (p + 1) % n; for (int i = 0; i < n; i++) { // If i is more counterclockwise than // current q, then update q if (orientation(points[p], points[i], points[q]) == 2) q = i; } // Now q is the most counterclockwise with // respect to p. Set p as q for next iteration, // so that q is added to result ‘hull‘ p = q; } while (p != l); // While we don‘t come to first // point // Print Result for (Point temp : hull) System.out.println("(" + temp.x + ", " + temp.y + ")"); } /* Driver program to test above function */ public static void main(String[] args) { Point points[] = new Point[7]; points[0]=new Point(0, 3); points[1]=new Point(2, 3); points[2]=new Point(1, 1); points[3]=new Point(2, 1); points[4]=new Point(3, 0); points[5]=new Point(0, 0); points[6]=new Point(3, 3); int n = points.length; convexHull(points, n); } }
标签:nbsp output other arc cte reac stat run sys
原文地址:http://www.cnblogs.com/apanda009/p/7965365.html