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;
}