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.....