Question

我有未完成的代码来找到所有垂直线的交点。到目前为止，我有这个：

import java.util.Scanner;

public class CountSquares {
    public static void main(String args[]) {
        Scanner scan = new Scanner(System.in);
        int lines = scan.nextInt();
        scan.nextLine();
        double[][] lineMXYC = new double[4][lines]; // stores the slope, and x
                                                    // and y co-ordinates and c
                                                    // so the line can be
                                                    // represented as y = mx + c
        double x1 = 0.0, x2 = 0.0, y1 = 0.0, y2 = 0.0;
        double slope = 0.0;
        for (int i = 0; i < lines; i++) {
            String input = scan.nextLine();
            String[] arrayOfInput = input.split(" ");
            x1 = Integer.parseInt(arrayOfInput[0]);
            y1 = Integer.parseInt(arrayOfInput[1]);
            x2 = Integer.parseInt(arrayOfInput[2]);
            y2 = Integer.parseInt(arrayOfInput[3]);
            if (x1 == x2)
                slope = Double.POSITIVE_INFINITY;
            else
                slope = (y2 - y1) / (x2 - x1);
            lineMXYC[0][i] = slope;
            lineMXYC[1][i] = x1;
            lineMXYC[2][i] = y1;
            lineMXYC[3][i] = y1 - (slope * x1);

        }

        for (int j = 0; j < lines - 1; j++) {  //sorts the array by slopes
            if (lineMXYC[0][j] > lineMXYC[0][j + 1]) {
                double TEMP = lineMXYC[0][j + 1];
                lineMXYC[0][j + 1] = lineMXYC[0][j];
                lineMXYC[0][j] = TEMP;
            }
        }
        double[] pointsOfIntersectionX = new double[(int) (lines * lines / 4) + 1]; //max number of points there can be
        double[] pointsOfIntersectionY = new double[(int) (lines * lines / 4) + 1];

        int count = 0;
        for (int k = 0; k < lines; k++) {
            for (int n = k; n < lines; n++) {
                System.out.println(n + " " + k);
                if (1 / lineMXYC[0][k] == -lineMXYC[0][n]) {
                    double m1 = lineMXYC[0][k];
                    double m2 = lineMXYC[0][n];
                    double c1 = lineMXYC[3][k];
                    double c2 = lineMXYC[3][n];
                    System.out.println("m1: "+m1);
                    System.out.println("m2: "+m2);
                    System.out.println("c1: "+c1);
                    System.out.println("c2: "+c2);
                    pointsOfIntersectionX[count] = (c1 - c2) / (m2 - m1); //determinate to find x co-ordinate
                    pointsOfIntersectionY[count] = (m2 * c1 - m1 * c2)
                            / (m2 - m1);
                    System.out.println("The lines intersect at: ("
                            + pointsOfIntersectionX[count] + ", "
                            + pointsOfIntersectionY[count] + ")");
                    count++;
                }
            }
        }

        scan.close();
    }
}

这将占用行数，然后每行以空格分隔两个点。如果我输入

2
3 -1 0 8
3 -1 0 -2

它工作正常，找到点（3，-1）但是，如果我把相同的值反转

2
3 -1 0 -2
3 -1 0 8

它给了我（-3.0,6.999999999999999）这是错误的

代码出了什么问题？我无法确定问题的来源。

Answer 1

按斜率排序可能会混合您的版本中的输入数据。如果要交换行，请交换所有行数据：

    for (int j = 0; j < lines - 1; j++) {  //sorts the array by slopes
        if (lineMXYC[0][j] > lineMXYC[0][j + 1]) {
            for (int i = 0; i < 4; i++) {
                double TEMP = lineMXYC[i][j + 1];
                lineMXYC[i][j + 1] = lineMXYC[i][j];
                lineMXYC[i][j] = TEMP;
            }
        }
    }

如果我可以给你一个建议：

为更好的设计制作单独的Line类
实施比较方法或行比较器
使用一些排序算法，您使用的算法将无法正确排序以获得更多行，例如：见Java Tutorials

确定不起作用的交叉点

1 个答案: