凸包练习: POJ 2187(JAVA)

分治化求凸包,请参看: http://128kj.iteye.com/admin/blogs/1748622
POJ 2187题意:
   给出一个点集，求两点之间最长的距离的平方，最长距离的两个点一定在凸包上，
首先，将点集凸包化，这样就可以排除了很多点，接下来就是两个for就可以.

下面是AC过代码:

import java.util.*;      
class Line {//线      
 Point p1, p2;      
 Line(Point p1, Point p2) {      
  this.p1 = p1;      
  this.p2 = p2;      
 }      
  public Point getP1(){
    return p1;
  }

  public Point getP2(){
    return 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;//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++) {   
              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);   
                } else if (area < 0) {   
                  ptsRight.add(p3);   
                }   
              }   
              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;   
    }   
}   
  
public class Main {   
   //计算距离的平方
  private static double dis(Point p1,Point p2){
     double dx = Math.abs(p1.x - p2.x);   
     double dy = Math.abs(p1.y - p2.y);   
     return (dx * dx + dy * dy);   
  }

  //计算最长距离的平方   
  private static void answer(List<Line> ls) {  
   
    List<Point> l=new ArrayList<Point>();//凸包中的所有点
   for (int i = 0; i < ls.size(); i++) {   
      Line line=ls.get(i);
      Point p1=line.getP1();
      Point p2=line.getP2();
      if(!l.contains(p1)) l.add(p1);
      if(!l.contains(p2)) l.add(p2);
   }   
   double ans=0;
    for (int i=0;i<l.size()-1;i++)
        for (int j=i+1;j<l.size();j++){
           double tem=dis(l.get(i),l.get(j));
            if (ans<tem)
                ans=tem;
        }
    System.out.printf("%.0f\n",ans);
    
  }   
  
  public static void main(String[] args) {   
    Scanner cin = new Scanner(System.in);   
    int n = cin.nextInt();   
    List<Point> pts = new ArrayList<Point>(n);   
    int x, y;   
    for (int i = 0; i < n; i++) {   
      x = cin.nextInt();   
      y = cin.nextInt();   
      pts.add(new Point(x, y));   
    }   
    QuickTuBao qt = new QuickTuBao(pts);   
    List<Line> ls = qt.eval();   
    answer(ls);
  }   
}