第六章第三十九题（几何：点的位置）(Geometry: point position)

第六章第三十九题（几何：点的位置）(Geometry: point position)

• 6.39（几何：点的位置）编程练习题3.32显示如何测试一个点是否在一个有向直线的左侧、右侧，或在该直线上。使用下面的方法头编写该方法：
  public static boolean leftOfTheLine(double x0, double y0, double x1, double y1, double x2, double y2)
  public static boolean onTheSameLine(double x0, double y0, double x1, double y1, double x2, double y2)
  public static boolean onTheLineSegment(double x0, double y0, double x1, double y1, double x2, double y2)
  编写一个程序，提示用户输入三个点赋给p0、p1和p2，显示p2是否在从p0到p1的直线的左侧、右侧、直线上，或者线段上。
  下面是一些运行示例：
  Enter three points for p0, p1, and p2: 1 1 2 2 1.5 1.5
  (1.5, 1.5) is on the line segment from (1.0, 1.0) to (2.0, 2.0)
  Enter three points for p0, p1, and p2: 1 1 2 2 3 3
  (3.0, 3.0) is on the same line from (1.0, 1.0) to (2.0, 2.0)
  Enter three points for p0, p1, and p2: 1 1 2 2 1 1.5
  (1.0, 1.5) is on the left side of the line from (1.0, 1.0) to (2.0, 2.0)
  Enter three points for p0, p1, and p2: 1 1 2 2 1 –1
  (1.0, −1.0) is on the right side of the line from (1.0, 1.0) to (2.0, 2.0)
  6.39(Geometry: point position) Programming Exercise 3.32 shows how to test whether a point is on the left side of a directed line, on the right, or on the same line. Write the methods with the following headers:
  public static boolean leftOfTheLine(double x0, double y0, double x1, double y1, double x2, double y2)
  public static boolean onTheSameLine(double x0, double y0, double x1, double y1, double x2, double y2)
  public static boolean onTheLineSegment(double x0, double y0, double x1, double y1, double x2, double y2)
  Write a program that prompts the user to enter the three points for p0, p1, and p2and displays whether p2 is on the left side of the line from p0 to p1, right side, the same line, or on the line segment.
  Here are some sample runs:
  Enter three points for p0, p1, and p2: 1 1 2 2 1.5 1.5
  (1.5, 1.5) is on the line segment from (1.0, 1.0) to (2.0, 2.0)
  Enter three points for p0, p1, and p2: 1 1 2 2 3 3
  (3.0, 3.0) is on the same line from (1.0, 1.0) to (2.0, 2.0)
  Enter three points for p0, p1, and p2: 1 1 2 2 1 1.5
  (1.0, 1.5) is on the left side of the line from (1.0, 1.0) to (2.0, 2.0)
  Enter three points for p0, p1, and p2: 1 1 2 2 1 –1
  (1.0, −1.0) is on the right side of the line from (1.0, 1.0) to (2.0, 2.0)
• 参考代码：

package chapter06;

import java.util.Scanner;

public class Code_39 {
    
    
    public static void main(String[] args) {
    
    
        Scanner inputScanner = new Scanner(System.in);
        System.out.print("Enter three points for p0, p1, and p2: ");
        double x0 = inputScanner.nextDouble();
        double y0 = inputScanner.nextDouble();
        double x1 = inputScanner.nextDouble();
        double y1 = inputScanner.nextDouble();
        double x2 = inputScanner.nextDouble();
        double y2 = inputScanner.nextDouble();
        if(onTheSameLine(x0, y0, x1, y1, x2, y2)) {
    
    
            if(onTheLineSegment(x0, y0, x1, y1, x2, y2)) {
    
    
                System.out.printf("(%f, %f) is on the line segment from " + "(%f, %f) to (%f, %f)", x2, y2, x0, y0, x1, y1);
            }
            else {
    
    
                System.out.printf("(%f, %f) is on the same line from " + "(%f, %f) to (%f, %f)", x2, y2, x0, y0, x1, y1);
            }
        }
        else {
    
    
            if(leftOfTheLine(x0, y0, x1, y1, x2, y2)) {
    
    
                System.out.printf("(%f, %f) is on the left side of the line from " + "(%f, %f) to (%f, %f)", x2, y2, x0, y0, x1, y1);
            }
            else {
    
    
                System.out.printf("(%f, %f) is on the right side of the line from " + "(%f, %f) to (%f, %f)", x2, y2, x0, y0, x1, y1);
            }
        }
    }
    public static boolean leftOfTheLine(double x0, double y0, double x1, double y1, double x2, double y2) {
    
    
        double discriminant = (x1 - x0) * (y2 - y0) - (x2 - x0) * (y1 - y0);

        if(discriminant > 0)
            return true;
        else
            return false;
    }
    public static boolean onTheSameLine(double x0, double y0, double x1, double y1, double x2, double y2) {
    
    
        double discriminant = (x1 - x0) * (y2 - y0) - (x2 - x0) * (y1 - y0);
        if(discriminant == 0)
            return true;
        else
            return false;
    }
    public static boolean onTheLineSegment(double x0, double y0, double x1, double y1, double x2, double y2) {
    
    
        double discriminant = (x1 - x0) * (y2 - y0) - (x2 - x0) * (y1 - y0);
        if(0 == discriminant && x2 <= (x0 > x1?x0:x1) && x2 >= (x0 > x1?x1:x0))
            return true;
        else
            return false;
    }
}

• 结果显示：

Enter three points for p0, p1, and p2: 1 1 2 2 1.5 1.5
(1.500000, 1.500000) is on the line segment from (1.000000, 1.000000) to (2.000000, 2.000000)
Process finished with exit code 0