学习凸包(二）:分治法求解
接上文:学习凸包(一）:暴力算法求解http://128kj.iteye.com/blog/1748442
import java.util.*; class Line {//线 Point p1, p2; Line(Point p1, Point p2) { this.p1 = p1; this.p2 = p2; } public double getLength() {double dx = Math.abs(p1.x - p2.x);double dy = Math.abs(p1.y - p2.y);return Math.sqrt(dx * dx + dy * dy); }} class Point{//点 double x; double y; public Point(double x,double y){ this.x=x; this.y=y; }} /**分治法求凸包*/class QuickTuBao { List<Point> pts = null;//给出的点集 List<Line> lines = new ArrayList<Line>();//点集pts的凸包 public void setPointList(List<Point> pts) { this.pts = pts; } public QuickTuBao(List<Point> pts){ this.pts=pts; } //求凸包，结果存入lines中 public List<Line> eval() { lines.clear(); if (pts == null || pts.isEmpty()) { return lines; } List<Point> ptsLeft = new ArrayList<Point>();//左凸包中的点 List<Point> ptsRight = new ArrayList<Point>();//右凸包中的点 //按x坐标对pts排序 Collections.sort(pts, new Comparator<Point>() { public int compare(Point p1, Point p2) { if(p1.x-p2.x>0) return 1; if(p1.x-p2.x<0) return -1; return 0; } }); Point p1 = pts.get(0);//最左边的点 Point p2 = pts.get(pts.size()-1);//最右边的点,用直线p1p2将原凸包分成两个小凸包 Point p3 = null; double area = 0; for (int i = 1; i < pts.size(); i++) {//穷举所有的点, p3 = pts.get(i); area = getArea(p1, p2, p3);//求此三点所成三角形的有向面积 if (area > 0) { ptsLeft.add(p3);//p3属于左 } else if (area < 0) { ptsRight.add(p3);//p3属于右 } } d(p1, p2, ptsLeft);//分别求解 d(p2, p1, ptsRight); return lines; } private void d(Point p1, Point p2, List<Point> s) { //s集合为空 if (s.isEmpty()) { lines.add(new Line(p1, p2)); return; } //s集合不为空，寻找Pmax double area = 0; double maxArea = 0; Point pMax = null; for (int i = 0; i < s.size(); i++) { area = getArea(p1, p2, s.get(i));//最大面积对应的点就是Pmax if (area > maxArea) { pMax = s.get(i); maxArea = area; } } //找出位于(p1, pMax)直线左边的点集s1 //找出位于(pMax, p2)直线左边的点集s2 List<Point> s1 = new ArrayList<Point>(); List<Point> s2 = new ArrayList<Point>(); Point p3 = null; for (int i = 0; i < s.size(); i++) { p3 = s.get(i); if (getArea(p1, pMax, p3) > 0) { s1.add(p3); } else if (getArea(pMax, p2, p3) > 0) { s2.add(p3); } } //递归 d(p1, pMax, s1); d(pMax, p2, s2); }// 三角形的面积等于返回值绝对值的二分之一// 当且仅当点p3位于直线(p1, p2)左侧时，表达式的符号为正private double getArea(Point p1, Point p2, Point p3) { return p1.x * p2.y + p3.x * p1.y + p2.x * p3.y - p3.x * p2.y - p2.x * p1.y - p1.x * p3.y;}}代码来自:http://blog.163.com/maxupeng@yeah/
未完,待续.....