Question

我已经在这几个小时，尝试不同的方法看几乎每个问题。也许我完全错了，但我觉得我的数学是正确的，但无论我输入什么数字，我都得到相同的输出。我的代码在某个地方关闭，我必须在午夜之前将其打开。

这一切都很有趣：找出一个点是否在三角形代码中。（适合初学者）

import java.util.Scanner;

public class PointsTriangle {

    // checks if point entered is within the triangle
    //given points of triangle are (0,0) (0,100) (200,0)
    public static void main (String [] args) {
        //obtain point (x,y) from user
        System.out.print("Enter a point's x- and y-coordinates: ");
        Scanner input = new Scanner(System.in);
        double x = input.nextDouble();
        double y = input.nextDouble();

        //find area of triangle with given points
        double ABC = ((0*(100-0  )+0*(0  -0)+200*(0-100))/2.0);
        double PAB = ((x*(0  -100)+0*(100-y)+0  *(y-  0))/2.0);
        double PBC = ((x*(100-0  )+0*(0  -y)+200*(y-100))/2.0);
        double PAC = ((x*(0  -100)+0*(100-y)+200*(y-  0))/2.0);

        boolean isInTriangle = PAB + PBC + PAC == ABC;

        if (isInTriangle)
            System.out.println("The point is in the triangle");
        else
            System.out.println("The point is not in the triangle");
    }//end main
}//end PointsTriangle

Answer 1

如果您绘制图片，您可以看到该点必须满足简单的不等式（某些线的右下方/上方/右侧）。无论“处于边缘”是否进入，我都会留给你：

Y > 0 (above the X axis)
X > 0 (to the right of the Y axis)
X + 2* Y < 200 (below the hypotenuse)

围绕这三个写一个if语句，你就完成了：

if( (y > 0) && (x > 0) && (x + 2*y < 200) ) 
  System.out.println("The point is in the triangle");
else
  System.out.println("The point is not in the triangle");

Answer 2

您将错误的值顺序放入公式中;因此，结果是错误的。如果3个顶点如下

A(x1, y1) B(x2, y2), C(x3, y3)

然后该区域计算为

double ABC = Math.abs (x1 * (y2 - y3) + x2 * (y3 - y1) + x3 * (y1 - y2)) / 2;

之后，你只需用输入点替换每个顶点，我们将有以下三角形：PBC，APC，ABP。

把所有东西放在一起，我们会有正确的

int x1 = 0, y1 = 0;
int x2 = 0, y2 = 100;
int x3 = 200, y3 = 0;

// no need to divide by 2.0 here, since it is not necessary in the equation
double ABC = Math.abs (x1 * (y2 - y3) + x2 * (y3 - y1) + x3 * (y1 - y2));
double ABP = Math.abs (x1 * (y2 - y) + x2 * (y - y1) + x * (y1 - y2));
double APC = Math.abs (x1 * (y - y3) + x * (y3 - y1) + x3 * (y1 - y));
double PBC = Math.abs (x * (y2 - y3) + x2 * (y3 - y) + x3 * (y - y2));

boolean isInTriangle = ABP + APC + PBC == ABC;

找出一个点是否在三角形内

2 个答案: