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C#折线是自交叉

[英]C# Polyline is self crossing

I've got a task to check if one polyline is self-crossing at any time. 我有一项任务可以随时检查一条折线是否自交。 This check must be very fast because my polyline is long (have about 50 points) and I've got a timeout. 此检查必须非常快,因为我的折线很长(大约有50点)并且我有超时。 Here is what I wrote: 这是我写的:

    public bool IsSelfCrossing()
    {
        if (size <= 5)
            return false;
        Point first = body.Points.ElementAt(size - 1);
        Point second = body.Points.ElementAt(size - 2);
        for (int i = 0; i < size - 3; i++)
        {
            if (Intersect(first, second, body.Points.ElementAt(i),
                body.Points.ElementAt(i + 1)))
            {
                return true;
            }
        }
        return false;
    }

    private double Orientation(Point p1, Point p2, Point p3)
    {
        double dx1 = p2.X - p1.X;
        double dy1 = p2.Y - p1.Y;
        double dx2 = p3.X - p1.X;
        double dy2 = p3.Y - p1.Y;
        return dx1 * dy2 - dy1 * dx2;
    }


    bool Intersect(Point p1, Point p2, Point p3, Point p4)
    {
        return
              Orientation(p1, p3, p4) * Orientation(p2, p3, p4) < 0 &&
              Orientation(p3, p1, p2) * Orientation(p4, p1, p2) < 0;
    }

The problem of these methods is that sometimes it fails (the methods are telling me that the polyline is self-crossing but it's not). 这些方法的问题是有时会失败(这些方法告诉我折线是自交叉的,但不是)。 Can you help me with better solution, please? 您能帮我提供更好的解决方案吗?

This paper describes sweep-line algorithm for finding intersections in set of line segments. 本文介绍了用于在线段集中查找相交的扫线算法。 It has expected running time of O(n + k) where n is number of segments and k is number of intersections. 它的预期运行时间为O(n + k),其中n是路段数,k是路口数。

http://www.cs.tufts.edu/comp/163/notes05/seg_intersection_handout.pdf http://www.cs.tufts.edu/comp/163/notes05/seg_intersection_handout.pdf

Here is a better implementation of your "Orientation" function, avoiding problems with rounding errors. 这是“方向”功能的更好实现,避免了舍入错误的问题。 Perhaps this helps in your case. 也许这对您有帮助。 It returns 0 if p0 is on a straight line between p1 and p2. 如果p0在p1和p2之间的直线上,则返回0。

    public static int Clockwise (Point p0, Point p1, Point p2)
    {
        const double epsilon = 1e-13;

        double dx1 = p1.X - p0.X;
        double dy1 = p1.Y - p0.Y;
        double dx2 = p2.X - p0.X;
        double dy2 = p2.Y - p0.Y;
        double d = dx1 * dy2 - dy1 * dx2;
        if(d > epsilon) return 1;
        if(d < -epsilon) return -1;
        if((dx1*dx2 < -epsilon) || (dy1*dy2 < -epsilon)) return -1;
        if((dx1*dx1+dy1*dy1) < (dx2*dx2+dy2*dy2)+epsilon) return 1;
        return 0;
    }

And here is my "Intersect" function: 这是我的“相交”功能:

    public static bool LineIntersects(Point p1,Point p2, Point q1,Point q2)
    {
        return (Clockwise(p1,p2,q1) * Clockwise(p1,p2,q2) <=0) &&
            (Clockwise(q1,q2,p1) * Clockwise(q1,q2,p2) <=0);
    }

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