Question

问题：我需要在C＃/ VB.NET中绘制如下图片
请注意，我的问题不是在C＃中绘制圆圈。 Ishihara transformation

我的问题是在没有太多空白和没有交叉的情况下绘制它们。
我的想法是使用“轨道”，然后绘制中心在“轨道”线上的圆圈。（有些轨道线用于较大的轨道线，有些仅用于较小的圆圈）问题是不应该有交叉点。

Orbit

任何人都有更好的主意？或者我如何测试给定半径的绘制圆是否与已存在的圆相交？

Answer 1

圆的交点很容易计算：如果沿x的差的平方加上沿y的差的平方小于半径之和的平方，则圆相交。

请注意，这已经进行了一些优化，因为它避免了取平方根。可以进行额外的优化，例如当沿x的差值大于半径之和时，它们将永远不会相交。

只需针对所有现有圈子测试新圈子，您就完成了。

这是O（n ^ 2），但是很容易且非常快，因为每次测试只是一些快速操作。

当然，您可以寻找一种优化，您不必对所有其他圈子测试每个圈子，但这些是昂贵的，大量的代码，因此只有很多圈子值得。首先尝试简单的解决方案。

在C ++代码中（抱歉，我不会说VB）：

struct { double x, y, r; } Circle;

bool circleIsAllowed(const std::vector<Circle>& circles, const Circle& newCircle)
{
   for(std::vector<Circle>::const_iterator it = circles.begin(); it != circles.end(); ++it) // foreach(Circle it in circles)
   {
      double sumR = it->r + newCircle.r; // + minimumDistanceBetweenCircles (if you want)
      double dx = it->x - newCircle.x;
      double dy = it->y - newCircle.y;
      double squaredDist = dx*dx + dy*dy;
      if (squaredDist < sumR*sumR) return false;
   }

   return true; // no existing circle overlaps
}

编辑：纠正了小错误，并注意到问题不是关于C ++

Answer 2

这是我尝试解释@ Sjoerd的代码（在VB.Net中）。代码来自标准的空白WinForm应用程序。它将一堆圆圈绘制成一个矩形。我将把它留给OP将这些限制在一个圆圈内。 DoCirclesIntersect函数采用可选的PadCircle参数，该参数尝试在圆圈之间提供更多间距，以便它们不会相互碰撞。它有点天真，但它似乎工作。你绘制的圆圈越多，它就越需要检查越来越多的边界。

Option Explicit On
Option Strict On

Public Class Form1
    ''//NOTE: The circles in this code are bound to a rectangle but it should be fairly trivial to create a master circle and check that

    ''//Dimension of the bounding image
    Private Shared ReadOnly ImageMaxDimension As Integer = 500
    Private Shared ReadOnly MinCircleDiameter As Integer = 4
    Private Shared ReadOnly MaxCircleDiameter As Integer = 15
    Private Shared ReadOnly CircleCount As Integer = 500
    Private Sub Form1_Load(ByVal sender As System.Object, ByVal e As System.EventArgs) Handles MyBase.Load
        ''//Create a picture box to output to
        Dim PB As New PictureBox()
        PB.Dock = DockStyle.Fill
        Me.Controls.Add(PB)

        ''//List of bounds of all circles created so far
        Dim AllBounds As New List(Of RectangleF)

        ''//Our random number generator
        Dim R As New Random()

        ''//Values for our individual circles
        Dim W, X, Y As Integer
        Dim Re As RectangleF

        ''//Create a bitmap to draw on
        Dim TempB As New Bitmap(ImageMaxDimension, ImageMaxDimension)
        Using G = Graphics.FromImage(TempB)
            For I = 1 To CircleCount
                ''//We can only draw so many circles, this just gives us a counter so we know when we reach the limit for a given size
                Trace.WriteLine(I)

                ''//Create an infinite loop that we will break out of if we have found a circle that does not intersect anything
                Do While True
                    ''//Create a random diameter
                    W = R.Next(MinCircleDiameter, MaxCircleDiameter + 1)
                    ''//Create a random X,Y
                    X = R.Next(0 + W, ImageMaxDimension - W)
                    Y = R.Next(0 + W, ImageMaxDimension - W)
                    ''//Create our rectangle
                    Re = New RectangleF(X, Y, W, W)

                    ''//Check each existing bound to see if they intersect with the current rectangle
                    For Each B In AllBounds
                        ''//If they do, start the loop over again
                        If DoCirclesIntersect(B, Re, 1) Then Continue Do
                    Next

                    ''//If we are here, no circles intersected, break from the infinite loop
                    Exit Do
                Loop

                ''//All the circle to our list
                AllBounds.Add(Re)

                ''/Draw the circle on the screen
                G.FillEllipse(Brushes.BurlyWood, Re)
            Next

            ''//Draw the image to the picture box
            PB.Image = TempB
        End Using
    End Sub
    Private Shared Function DoCirclesIntersect(ByVal r1 As RectangleF, ByVal r2 As RectangleF, Optional ByVal PadCircle As Integer = 0) As Boolean
        ''//This code is hopefully what @Sjoerd said in his post
        Dim aX = Math.Pow(r1.X - r2.X, 2)
        Dim aY = Math.Pow(r1.Y - r2.Y, 2)
        Dim Dif = Math.Abs(aX - aY)
        Dim ra1 = r1.Width / 2
        Dim ra2 = r2.Width / 2
        Dim raDif = Math.Pow(ra1 + ra2, 2)
        Return (raDif + PadCircle) > Dif
    End Function
End Class

如何绘制ishihara变换（圆圈没有交叉）？

2 个答案: