Convex hull - Divide and Conquer
Request can be entirely contained on a plane to the n-th point of a convex polygon.
Example:
First, divide and conquer:
(A) algorithm ideas:
(Mentioned here are all straight line to a straight line.)
The array in ascending order, if the same x coordinate, y-axis coordinates sorted in ascending order.
Leftmost point p1 and the rightmost point p_n must convex hull of the set of vertices. The straight line is divided into two set points, the package as S1, the next packet to S2. Impossible to point p1 p_n line is a convex hull of the vertices, so we do not consider.
S1 in the package, find P_max (the most distant point from the straight line p1p_n), if two points far from the same, to take maximum ∠p_max p1 p_n that point (i.e., △ p_max p1 p_n largest area).
(Recursive ends here)
Find all points in S1 to the left of the straight line p1 P_max, some of these points constituting the vertices on the left half of the packet boundary, with the above algorithm recursive search point until the pack is to the endpoint p1 and p_n segment.
Under the package S2 in the same way to find the next border.
* How to determine whether the point in a straight line p1 left p_max (with p1 p_n above)?
If q1 (x1, y1), q2 (x2, y2), q3 (x3, y3) is arbitrary three points in the plane, then the triangle △ q1 q2 q3 following area equal to one-half the absolute value of determinant .
If and only if the point q3 = (x3, y3) on the left side of the straight line q1 q2, the symbolic expression is positive, the point is determined by two points on the left side of the line.
(B) implementation problems encountered
How to quickly sort to sort the Point class (there are coordinates x, y within) a one-dimensional array?
According to the x coordinate ordering is very simple, if encountered the same x, y, how different do?
In the original basis of the fast row modified to forward j Approach Description:
When (first while loop) the current number of comparison abscissa> reference point, j forward approximation. Here without equal numbers, sorting is unstable, i.e., the relative position of the equal element may change. (See blog fast row: https: //www.cnblogs.com/musecho/p/11647349.html)
(Second while add content) Comparative ordinate equal elements, a reference point is smaller, j approximation continue forward, that is equal to the relative position of the element does not change; otherwise, is changed. That is the original row fast while loop is split into two, equal elements increase further the comparative ordinate.
while (i < j && points[j].getX() > center.getX()) {
j--;
}
while (i < j && center.getX() == points[j].getX() && points[j].getY() > center.getY()) {
j--;
}
/*
* (i<j)若points[j].getX()< center.getX()或 center.getX() ==
* points[j].getX()且points[j].getY()<center.getY() 以上两种情况，需要赋值
*/
if (i < j)// 跳出循环也有可能时因为i=j，所以这里要判断一下
points[i++] = points[j];- If you are using global arrays visit to identify whether the access point can determine all the vertices of the convex hull, but how the sequence output?
Has been seeking in the apex of the convex hull boundary is determined individually, is determined that not all points in this side of the border, if it is determined that a border.
convexHullList.add(convexHullVertex[0]);// 开始点
int haveCount = 1;// 已经加入点的个数
// 逐条确定边界，判断是否除了该条假设边界上的点，其他凸包的顶点都在直线的右边。
// 如果是，则此条直线为边界；如果不是，取下一个边界终点，继续判断。
int start = 0;// 起点
for (int end = start + 1; haveCount < count;) {
boolean boundRight = true;
for (int i = 0; i < count; i++) {
while (i < count && (i == start || i == end)) {// 不能写if，start和end可能是连在一起的
i++;
}
if (i >= count)
break;
// 点在直线左侧或线上，错误
if (PointJudge(convexHullVertex[start], convexHullVertex[end], convexHullVertex[i]) >= 0) {
boundRight = false;
end = (end + 1) % count;// end取下一个
break;
}
}
if (boundRight == true) {
convexHullList.add(convexHullVertex[end]);
start = end;
end = (start + 1) % count;
haveCount++;
}
}(C) Precautions
Note Method PointJudge (Point beginP, Point endP, Point p) and PointCal (Point beginP, Point endP, Point p), the parameter passing on the first few:
The first two points are the two ends of a straight line, the third point is the need to determine
- Note that the packet cycle starting and end points, the determination condition
for (int i = begin - 1; i >= end + 1; i--)(D) source code
1.ConvexHullProblem_DC
package ConvexHullProblem;
import java.util.ArrayList;
import java.util.Arrays;
/**
* 凸包问题（分治法）：
*/
public class ConvexHullProblem_DC {
boolean[] visit;// 标志点是否是凸包的顶点：1是，0不是
Point[] points;// 所有点
Point[] convexHullVertex;// 凸包的顶点
ArrayList<Point> convexHullList = new ArrayList<>();;// 凸包的顶点（顺序存放）
public void ConvexHullProblem(Point[] points) {
this.points = points;
quickSort(0, points.length - 1);// 升序排序
// System.out.println("升序：" + Arrays.toString(points));
visit = new boolean[points.length];
recursion(0, points.length - 1);// 上包
recursion(points.length - 1, 0);// 下包
orderConvexHull();
}
/**
* @title: recursion
* @description: 在凸包的上包或下包中，找使△p_max p1 p_n面积最大的点p_max，并递归
* @date: 2019年10月16日 下午8:33:14
* @param begin 直线的起点
* @param end 直线的终点 void
* @throws:
*/
void recursion(int begin, int end) {
// 直线的两端点为凸包的顶点
visit[begin] = true;
visit[end] = true;
if (begin < end) {
boolean flag = false;// 标志直线左侧是否有点
int maxArea = 0;// 最大面积
int maxAreaindex = begin + 1;// 最大面积的点下标
for (int i = begin + 1; i <= end - 1; i++) {// begin和end已经是顶点，不需要判断
if (PointJudge(points[begin], points[end], points[i]) > 0) {// 点在直线左侧
// 找距离最远的点，因为底相同都是p1 pn，也就是求三角形面积最大的
flag = true;
int area = PointCal(points[begin], points[end], points[i]);
if (area > maxArea) {
maxArea = area;
maxAreaindex = i;
} else if (area == maxArea) {// 若面积相同，取∠p_max p_begin p_end最大的那一个
System.out.println(22);
double degreeA = Degree(points[begin], points[i], points[end]);
double degreeB = Degree(points[begin], points[maxAreaindex], points[end]);
if (degreeA > degreeB) {
maxArea = area;
maxAreaindex = i;
}
}
// System.out.println("area=" + area + ",Point=" + points[i]);
}
}
// System.out.println("maxArea=" + maxArea + ",Point=" + points[maxAreaindex]);
// System.out.println("over");
// 若直线左侧还有点，则递归；没有点，则结束
if (flag == true) {
recursion(begin, maxAreaindex);
recursion(maxAreaindex, end);
}
} else if (begin > end) {
boolean flag = false;
int maxArea = 0;// 最大面积
int maxAreaindex = end + 1;// 最大面积的点下标
for (int i = begin - 1; i >= end + 1; i--) {// 注意下包循环中的 起始点、终点、判断条件
if (PointJudge(points[begin], points[end], points[i]) > 0) {// 点在直线左侧
flag = true;
int area = PointCal(points[begin], points[end], points[i]);
if (area > maxArea) {
maxArea = area;
maxAreaindex = i;
} else if (area == maxArea) {// 若面积相同，取∠p_max p_begin p_end最大的那一个
System.out.println(22);
double degreeA = Degree(points[begin], points[i], points[end]);
double degreeB = Degree(points[begin], points[maxAreaindex], points[end]);
if (degreeA > degreeB) {
maxArea = area;
maxAreaindex = i;
}
}
// System.out.println("area=" + area + ",Point=" + points[i]);
}
}
// System.out.println("maxArea=" + maxArea + ",Point=" + points[maxAreaindex]);
// System.out.println("over");
if (flag == true) {
recursion(begin, maxAreaindex);
recursion(maxAreaindex, end);
}
}
}
/**
* @title: quickSort：运用Hoare
* @description: 快速排序： 选取第一个元素作为基准点（可以随机选取），将剩下元素与基准点进行比较，
* 比基准点大的放在右边，比基准点小的放在左边， 得到左子表和右子表，递归调用本函数；
* @param points 数组
* @param begin 开始下标
* @param end 结束下标
* @throws:
*/
void quickSort(int begin, int end) {
if (begin >= 0 && begin < end && end < points.length) {
int i = begin, j = end;
Point center = points[i];// 中心元素
while (i != j) {
while (i < j && points[j].getX() > center.getX()) {
j--;
}
while (i < j && center.getX() == points[j].getX() && points[j].getY() > center.getY()) {
j--;
}
/*
* (i<j)若points[j].getX()< center.getX()或 center.getX() ==
* points[j].getX()且points[j].getY()<center.getY() 以上两种情况，需要赋值
*/
if (i < j)// 跳出循环也有可能时因为i=j，所以这里要判断一下
points[i++] = points[j];
while (i < j && points[i].getX() < center.getX()) {
i++;
}
while (i < j && points[i].getX() == center.getX() && points[i].getY() < center.getY()) {
i++;
}
/*
* (i<j)若points[i].getX()> center.getX()或 center.getX() ==
* points[i].getX()且points[i].getY()>center.getY() 以上两种情况，需要赋值
*/
if (i < j)
points[j--] = points[i];
}
points[i] = center;// 中心元素到达最终位置
quickSort(begin, i - 1);
quickSort(i + 1, end);
}
}
/**
* @title: PointCal
* @description: 计算行列式的值
* @date: 2019年10月15日 下午7:53:07
* @param beginP 直线的开始点
* @param p 判断的点
* @param endP 直线的终点
* @return int 行列书的值
* @throws:
*/
private int PointCal(Point beginP, Point endP, Point p) {
int cal = 0;// 行列式值
//x1y2+x3y1+x2y3-x3y2-x2y1-x1y3
cal = beginP.getX() * endP.getY() + p.getX() * beginP.getY() + endP.getX() * p.getY() - p.getX() * endP.getY()
- endP.getX() * beginP.getY() - beginP.getX() * p.getY();
return cal;
}
/**
* @title: PointJudge
* @description:返回点p在直线beginP endP的位置
* @date: 2019年10月15日 下午7:56:56
* @param beginP
* @param p 判断的点
* @param endP
* @return int ：1在直线左侧，0在线上，-1在右侧
* @throws: 注意传参放在第几个，前两个点是直线的两端，第三个是需要判断的点
*/
private int PointJudge(Point beginP, Point endP, Point p) {
if (PointCal(beginP, endP, p) > 0) {
return 1;
} else if (PointCal(beginP, endP, p) == 0)
return 0;
else
return -1;
}
/**
* @title: Degree
* @description: 余弦公式求∠pa pb pc的度数
* @date: 2019年10月16日 下午6:59:29
* @param pa 点
* @param pb
* @param pc
* @return double：返回∠c的度数（°为单位）
* @throws:
*/
double Degree(Point pa, Point pb, Point pc) {
double degree = 0;// ∠pa pb pc度数
// 三角形的三边长
double a = Math.sqrt(Math.pow(pa.getX() - pb.getX(), 2) + Math.pow(pa.getY() - pb.getY(), 2));
double b = Math.sqrt(Math.pow(pb.getX() - pc.getX(), 2) + Math.pow(pb.getY() - pc.getY(), 2));
double c = Math.sqrt(Math.pow(pc.getX() - pa.getX(), 2) + Math.pow(pc.getY() - pa.getY(), 2));
// 余弦公式求∠pa pb pc度数
System.out.println("acos=" + Math.acos((a * a + b * b - c * c) / (2.0 * a * b)));
degree = Math.toDegrees(Math.acos((a * a + b * b - c * c) / (2.0 * a * b)));
System.out.println("degree=" + degree);
return degree;
}
/**
*@title: orderConvexHull
*@description: 凸包顶点按顺时针输出
*@date: 2019年10月19日 上午9:28:44
*void
*@throws:
*/
void orderConvexHull() {
/** 将凸包顶点存放进另一个数组 */
int count = 0;// 凸包的顶点个数
for (int i = 0; i < visit.length; i++) {
if (visit[i] == true) {
count++;
}
}
convexHullVertex = new Point[count];
for (int j = 0, i = 0; j < visit.length; j++) {
if (visit[j] == true) {
convexHullVertex[i] = points[j];
i++;
}
}
convexHullList.add(convexHullVertex[0]);// 开始点
int haveCount = 1;// 已经加入点的个数
// 逐条确定边界，判断是否除了该条假设边界上的点，其他凸包的顶点都在直线的右边。
// 如果是，则此条直线为边界；如果不是，取下一个边界终点，继续判断。
int start = 0;// 起点
for (int end = start + 1; haveCount < count;) {
boolean boundRight = true;
for (int i = 0; i < count; i++) {
while (i < count && (i == start || i == end)) {// 不能写if，start和end可能是连在一起的
i++;
}
if (i >= count)
break;
// 点在直线左侧或线上，错误
if (PointJudge(convexHullVertex[start], convexHullVertex[end], convexHullVertex[i]) >= 0) {
// System.out.println("****");
// System.out.println(convexHullVertex[start]);
// System.out.println(convexHullVertex[end]);
// System.out.println(convexHullVertex[i]);
// System.out.println("****");
boundRight = false;
end = (end + 1) % count;// end取下一个
break;
}
}
if (boundRight == true) {
convexHullList.add(convexHullVertex[end]);
start = end;
end = (start + 1) % count;
haveCount++;
}
}
convexHullList.add(convexHullVertex[0]);// 结束点
System.out.println("凸包顶点顺时针输出：" + convexHullList);
}
}
2.Point
package ConvexHullProblem;
/**
* 点的信息
*/
public class Point {
private int x, y;// 横纵坐标
public Point(int x,int y) {
this.x=x;
this.y=y;
}
public int getX() {
return x;
}
public void setX(int x) {
this.x = x;
}
public int getY() {
return y;
}
public void setY(int y) {
this.y = y;
}
public String toString() {
return " (" + x + ", "+ y + ")";
}
}
3.Demo_DC
package ConvexHullProblem;
import java.util.Arrays;
import java.util.Scanner;
public class Demo_DC {
public static void main(String[] args) {
//示例
Point[] points=new Point[13];
points[0] = new Point( 4, 5);
points[1] = new Point(10, 11);
points[2] = new Point( 4, 11);
points[3] = new Point( 1, 1);
points[4] = new Point( 10, 6);
points[5] = new Point( 8, 14);
//横坐标最两侧的点
points[6] = new Point( 13, 7);
points[7] = new Point( 13, 0);
points[8] = new Point( 0, 9);
points[9] = new Point( 0, 7);
points[10] = new Point( 5, 5);
points[11] = new Point( 7, 9);
points[12] = new Point( 11, 3);
// Scanner scanner = new Scanner(System.in);
// System.out.println("**********凸包问题**********");
// System.out.println("请输入点的数量：");
// int n = scanner.nextInt();
// Point[] points = new Point[n];
// points = randomPoint(points);
System.out.println("随机生成点：" + Arrays.toString(points));
ConvexHullProblem_DC convexHullProblem = new ConvexHullProblem_DC();
convexHullProblem.ConvexHullProblem(points);
}
static Point[] randomPoint(Point[] points) {
for (int i = 0; i < points.length; i++) {
int x = (int) (Math.random() * 21);// [0-20]
int y = (int) (Math.random() * 21);// [0-20]
points[i] = new Point(x, y);
}
return points;
}
}
(operation result)
