c# - 如何知道一条线是否与C#中的平面相交？

Question

我有两个点（一条线段）和一个矩形。我想知道如何计算线段是否与矩形相交。

Answer 1

从我的“几何”课：

public struct Line
{
    public static Line Empty;

    private PointF p1;
    private PointF p2;

    public Line(PointF p1, PointF p2)
    {
        this.p1 = p1;
        this.p2 = p2;
    }

    public PointF P1
    {
        get { return p1; }
        set { p1 = value; }
    }

    public PointF P2
    {
        get { return p2; }
        set { p2 = value; }
    }

    public float X1
    {
        get { return p1.X; }
        set { p1.X = value; }
    }

    public float X2
    {
        get { return p2.X; }
        set { p2.X = value; }
    }

    public float Y1
    {
        get { return p1.Y; }
        set { p1.Y = value; }
    }

    public float Y2
    {
        get { return p2.Y; }
        set { p2.Y = value; }
    }
}

public struct Polygon: IEnumerable<PointF>
{
    private PointF[] points;

    public Polygon(PointF[] points)
    {
        this.points = points;
    }

    public PointF[] Points
    {
        get { return points; }
        set { points = value; }
    }

    public int Length
    {
        get { return points.Length; }
    }

    public PointF this[int index]
    {
        get { return points[index]; }
        set { points[index] = value; }
    }

    public static implicit operator PointF[](Polygon polygon)
    {
        return polygon.points;
    }

    public static implicit operator Polygon(PointF[] points)
    {
        return new Polygon(points);
    }

    IEnumerator<PointF> IEnumerable<PointF>.GetEnumerator()
    {
        return (IEnumerator<PointF>)points.GetEnumerator();
    }

    public IEnumerator GetEnumerator()
    {
        return points.GetEnumerator();
    }
}

public enum Intersection
{
    None,
    Tangent,
    Intersection,
    Containment
}

public static class Geometry
{

    public static Intersection IntersectionOf(Line line, Polygon polygon)
    {
        if (polygon.Length == 0)
        {
            return Intersection.None;
        }
        if (polygon.Length == 1)
        {
            return IntersectionOf(polygon[0], line);
        }
        bool tangent = false;
        for (int index = 0; index < polygon.Length; index++)
        {
            int index2 = (index + 1)%polygon.Length;
            Intersection intersection = IntersectionOf(line, new Line(polygon[index], polygon[index2]));
            if (intersection == Intersection.Intersection)
            {
                return intersection;
            }
            if (intersection == Intersection.Tangent)
            {
                tangent = true;
            }
        }
        return tangent ? Intersection.Tangent : IntersectionOf(line.P1, polygon);
    }

    public static Intersection IntersectionOf(PointF point, Polygon polygon)
    {
        switch (polygon.Length)
        {
            case 0:
                return Intersection.None;
            case 1:
                if (polygon[0].X == point.X && polygon[0].Y == point.Y)
                {
                    return Intersection.Tangent;
                }
                else
                {
                    return Intersection.None;
                }
            case 2:
                return IntersectionOf(point, new Line(polygon[0], polygon[1]));
        }

        int counter = 0;
        int i;
        PointF p1;
        int n = polygon.Length;
        p1 = polygon[0];
        if (point == p1)
        {
            return Intersection.Tangent;
        }

        for (i = 1; i <= n; i++)
        {
            PointF p2 = polygon[i % n];
            if (point == p2)
            {
                return Intersection.Tangent;
            }
            if (point.Y > Math.Min(p1.Y, p2.Y))
            {
                if (point.Y <= Math.Max(p1.Y, p2.Y))
                {
                    if (point.X <= Math.Max(p1.X, p2.X))
                    {
                        if (p1.Y != p2.Y)
                        {
                            double xinters = (point.Y - p1.Y) * (p2.X - p1.X) / (p2.Y - p1.Y) + p1.X;
                            if (p1.X == p2.X || point.X <= xinters)
                                counter++;
                        }
                    }
                }
            }
            p1 = p2;
        }

        return (counter % 2 == 1) ? Intersection.Containment : Intersection.None;
    }

    public static Intersection IntersectionOf(PointF point, Line line)
    {
        float bottomY = Math.Min(line.Y1, line.Y2);
        float topY = Math.Max(line.Y1, line.Y2);
        bool heightIsRight = point.Y >= bottomY &&
                             point.Y <= topY;
        //Vertical line, slope is divideByZero error!
        if (line.X1 == line.X2)
        {
            if (point.X == line.X1 && heightIsRight)
            {
                return Intersection.Tangent;
            }
            else
            {
                return Intersection.None;
            }
        }
        float slope = (line.X2 - line.X1)/(line.Y2 - line.Y1);
        bool onLine = (line.Y1 - point.Y) == (slope*(line.X1 - point.X));
        if (onLine && heightIsRight)
        {
            return Intersection.Tangent;
        }
        else
        {
            return Intersection.None;
        }
    }

}

Answer 2

对矩形的直线和每一边执行http://mathworld.wolfram.com/Line-LineIntersection.html。
或： http: //mathworld.wolfram.com/Line-PlaneIntersection.html

Answer 3

因为它丢失了，我只是为了完整性而添加它

public static Intersection IntersectionOf(Line line1, Line line2)
    {
        //  Fail if either line segment is zero-length.
        if (line1.X1 == line1.X2 && line1.Y1 == line1.Y2 || line2.X1 == line2.X2 && line2.Y1 == line2.Y2)
            return Intersection.None;

        if (line1.X1 == line2.X1 && line1.Y1 == line2.Y1 || line1.X2 == line2.X1 && line1.Y2 == line2.Y1)
            return Intersection.Intersection;
        if (line1.X1 == line2.X2 && line1.Y1 == line2.Y2 || line1.X2 == line2.X2 && line1.Y2 == line2.Y2)
            return Intersection.Intersection;

        //  (1) Translate the system so that point A is on the origin.
        line1.X2 -= line1.X1; line1.Y2 -= line1.Y1;
        line2.X1 -= line1.X1; line2.Y1 -= line1.Y1;
        line2.X2 -= line1.X1; line2.Y2 -= line1.Y1;

        //  Discover the length of segment A-B.
        double distAB = Math.Sqrt(line1.X2 * line1.X2 + line1.Y2 * line1.Y2);

        //  (2) Rotate the system so that point B is on the positive X axis.
        double theCos = line1.X2 / distAB;
        double theSin = line1.Y2 / distAB;
        double newX = line2.X1 * theCos + line2.Y1 * theSin;
        line2.Y1 = line2.Y1 * theCos - line2.X1 * theSin; line2.X1 = newX;
        newX = line2.X2 * theCos + line2.Y2 * theSin;
        line2.Y2 = line2.Y2 * theCos - line2.X2 * theSin; line2.X2 = newX;

        //  Fail if segment C-D doesn't cross line A-B.
        if (line2.Y1 < 0 && line2.Y2 < 0 || line2.Y1 >= 0 && line2.Y2 >= 0)
            return Intersection.None;

        //  (3) Discover the position of the intersection point along line A-B.
        double posAB = line2.X2 + (line2.X1 - line2.X2) * line2.Y2 / (line2.Y2 - line2.Y1);

        //  Fail if segment C-D crosses line A-B outside of segment A-B.
        if (posAB < 0 || posAB > distAB)
            return Intersection.None;

        //  (4) Apply the discovered position to line A-B in the original coordinate system.
        return Intersection.Intersection;
    }

请注意，该方法会旋转线段以避免与方向相关的问题

Answer 4

如果是 2d，那么所有的线都在唯一的平面上。

所以，这是基本的 3-D 几何。你应该能够用一个简单的方程来做到这一点。

看看这个页面：

http://local.wasp.uwa.edu.au/~pbourke/geometry/planeline/。

第二种解决方案应该很容易实现，只要将矩形的坐标转换为平面方程即可。

此外，检查您的分母是否不为零（线不相交或包含在平面中）。

Answer 5

使用类：

System.Drawing.Rectangle

方法：

IntersectsWith();

Answer 6

我讨厌浏览 MSDN 文档（它们非常慢而且很奇怪 :-s），但我认为它们应该有类似于这种 Java 方法的东西……如果没有，对他们不利！XD（顺便说一句，它适用于段，而不是线条）。

无论如何，你可以看看开源 Java SDK 是如何实现的，也许你会学到一些新的技巧（当我看别人的代码时，我总是很惊讶）

Answer 7

是否可以使用简单的线段公式检查矩形每一侧的线。