1

所以我在 Windows 窗体应用程序中得到了一个圆圈，并且必须在这个圆圈中放置 20 个随机点。我的想法是将圆圈分成 4 部分以使其更加平衡。我的问题是这些点都是在中间产生的，我不知道如何解决这个问题......

    Graphics g;
    Pen p;
    Random r = new Random();
    int[] KegelX = new int[20];
    int[] KegelY = new int[20];
    private void Form1_Load(object sender, EventArgs e)
    {

    }

    private void button1_Click(object sender, EventArgs e)
    {
        Kegelplatzierung();
        p = new Pen(Color.Black);
        g = this.CreateGraphics();
        g.DrawEllipse(p, new Rectangle(50, 50, 400, 400));
        for (int i = 0; i < 20; i++)
        {
            g.DrawEllipse(p, new Rectangle(KegelX[i], KegelY[i], 1, 1));
        }
        p.Dispose();
        g.Dispose();
    }

    private void Kegelplatzierung() {
        for (int i = 0; i < 5; i++)
        {
            bool Kriterium = false;
            while (!Kriterium)
            {
                KegelX[i] = r.Next(50, 250);
                KegelY[i] = r.Next(50, 250);
                if (Math.Sqrt((250 - KegelX[i]) ^ 2 + (KegelY[i] - 250) ^ 2) < 200)
                {
                    Kriterium = true;
                }
            }
        }
        for (int i = 5; i < 10; i++)
        {
            bool Kriterium = false;
            while (!Kriterium)
            {
                KegelX[i] = r.Next(250, 450);
                KegelY[i] = r.Next(50, 250);
                if (Math.Sqrt((KegelX[i] - 250) ^ 2 + (KegelY[i] - 250) ^ 2) < 200)
                {
                    Kriterium = true;
                }
            }
        }
        for (int i = 10; i < 15; i++)
        {
            bool Kriterium = false;
            while (!Kriterium)
            {
                KegelX[i] = r.Next(50, 250);
                KegelY[i] = r.Next(250, 450);
                if (Math.Sqrt((250 - KegelX[i]) ^ 2 + (250 - KegelY[i]) ^ 2) < 200)
                {
                    Kriterium = true;
                }
            }
        }
        for (int i = 15; i < 20; i++)
        {
            bool Kriterium = false;
            while (!Kriterium)
            {
                KegelX[i] = r.Next(250, 450);
                KegelY[i] = r.Next(250, 450);
                if (Math.Sqrt((KegelX[i] - 250) ^ 2 + (250 - KegelY[i]) ^ 2) < 200)
                {
                    Kriterium = true;
                }
            }
        }

    }

示例： http: //puu.sh/gB6Dg/e81f8c3486.png http://puu.sh/gB6Ec/306f61424c.png

感谢帮助！

4

4 回答 4

3

这将按照您的计划进行。但是请注意，四个象限之间的对称性只有在您确实需要它来平衡很少的点时才是必要的。对于更大的数字，没有必要，您可以将代码减少到其行数的一半左右..！

private void panel2_Paint(object sender, PaintEventArgs e)
{
    int dotsPerQuadrant = 666;
    Random R = new Random();
    Size s1x1 = new System.Drawing.Size(2, 2);
    int radius = 200;
    int rad2 = radius / 2;
    int off = 20;
    Rectangle bounds = new Rectangle(off, off, radius, radius);
    GraphicsPath gp = new GraphicsPath();
    gp.AddEllipse(bounds);
    Rectangle rectQ1 = new Rectangle(off, off, rad2, rad2);
    Rectangle rectQ2 = new Rectangle(off + rad2, off, rad2, rad2);
    Rectangle rectQ3 = new Rectangle(off, off + rad2, rad2, rad2);
    Rectangle rectQ4 = new Rectangle(off + rad2, off + rad2, rad2, rad2);
    List<Rectangle> quadrants = new List<Rectangle> { rectQ1, rectQ2, rectQ3, rectQ4 };
    e.Graphics.Clear(Color.AntiqueWhite);
    e.Graphics.DrawEllipse(Pens.Black, bounds);

    foreach (Rectangle rect in quadrants)
    {
        int count = 0;
        do
        {
            Point p = new Point(rect.X + R.Next(rad2), rect.Y + R.Next(rad2));
            if (gp.IsVisible(p))
            {
                e.Graphics.FillEllipse(Brushes.Red, new Rectangle(p, s1x1));
                count++;
            }
        } while (count < dotsPerQuadrant);
    }

}

这是结果。点均匀地分布在整个圆圈上，而不是聚集在中间：

4x666 随机点

直接代码，无象限；

private void panel2_Paint(object sender, PaintEventArgs e)
{
    int dotstoDraw = 666*4;
    Random R = new Random();
    Size s1x1 = new System.Drawing.Size(2, 2);
    int radius = 200;
    int off = 20;
    Rectangle bounds = new Rectangle(off, off, radius, radius);
    GraphicsPath gp = new GraphicsPath();
    gp.AddEllipse(bounds);

    e.Graphics.Clear(Color.AntiqueWhite);
    e.Graphics.DrawEllipse(Pens.Black, bounds);

    int count = 0;
    do
    {
        Point p = new Point(off + R.Next(radius), off + R.Next(radius));
        if (gp.IsVisible(p))
        {
            e.Graphics.FillEllipse(Brushes.Red, new Rectangle(p, s1x1));
            count++;
        }
    } while (count < dotstoDraw);
 }
于 2015-03-15T13:28:57.287 回答
2

您可以通过为每个点选择随机半径和角度来做到这一点。

为了避免将角度分量明显量化为径向辐条，我使用(double)rand.Next() / int.MaxValue了一个介于 0 和 1 之间的随机数并将其乘以 2π。

为了避免点在圆心附近聚集，我使用了 Chris A. 的公式（在Generate a random point within a circle (uniformly)）来生成半径：

Random rand = new Random();

private List<Point> GetRandomPoints(double rMax, int nPoints)
{
    var randPoints = new List<Point>();
    for (int i = 0; i < nPoints; i++)
    {
        var r = Math.Sqrt((double)rand.Next() / int.MaxValue) * rMax;
        var theta = (double)rand.Next() / int.MaxValue * 2 * Math.PI;
        randPoints.Add(new Point((int)(r * Math.Cos(theta)), (int)(r * Math.Sin(theta))));
    }
    return randPoints;
}

private void button1_Click(object sender, EventArgs e)
{
    using (Graphics g = this.CreateGraphics())
    {
        using (Pen p = new Pen(Color.Black))
        {
            var left = 50;
            var top = 50;
            var r = 200;
            g.DrawEllipse(p, new Rectangle(left, top, r * 2, r * 2));

            int nPoints = 20;
            var randomPoints = GetRandomPoints(r - 1, nPoints);

            for (int i = 0; i < nPoints; i++)
            {
                g.DrawEllipse(p, new Rectangle(randomPoints[i].X + left + r, randomPoints[i].Y + top + r, 1, 1));
            }
        }
    }
}

为了确保对象被干净地处理，您可以使用该using构造 - 它可以确保您不会意外忘记。

通常最好将诸如圆的半径之类的东西分配给一个变量，因为这样您就可以在一个地方轻松地更改它，并且如果您使用有意义的变量名，它会使代码更易于阅读。

这是一个输出示例，但半径设置为 100 并生成 200 个点：

样本输出

于 2015-03-15T13:22:12.487 回答
2

问题是在 C# 中，^是逻辑异或运算符。您需要改用Math.Pow。所以...

if (Math.Sqrt(Math.Pow(250 - KegelX[i], 2) + Math.Pow(KegelY[i] - 250, 2)) < 200)

等等。

于 2015-03-15T13:46:48.447 回答
1

在给定半径的圆内生成一组随机点...

 private void button1_Click(object sender, EventArgs e)
            {
                int radius = 100;
                var p = new Pen(Color.Black);
                var g = this.CreateGraphics();
                g.DrawEllipse(p, 0,0,radius*2, radius*2);
                var pointGen = new RandomPointGenerator();
                var randomPoints = pointGen.GetPointsInACircle(radius, 20);
                p.Color = Color.Red;
                foreach (var point in randomPoints)
                {

                    g.DrawEllipse(p, point.X + radius, point.Y+radius, 2, 2);
                }
            }



public class RandomPointGenerator
    {
        private Random _randy = new Random();
        public List<Point> GetPointsInACircle(int radius, int numberOfPoints)
        {
            List<Point> points = new List<Point>();
            for (int pointIndex = 0; pointIndex < numberOfPoints; pointIndex++)
            {
                int distance = _randy.Next(radius);
                double angleInRadians = _randy.Next(360)/(2 * Math.PI) ;

                int x = (int)(distance * Math.Cos(angleInRadians));
                int y = (int)(distance * Math.Sin(angleInRadians));
                Point randomPoint = new Point(x, y);
                points.Add(randomPoint);
            }
            return points;
        }
    }
于 2015-03-15T13:06:54.053 回答