9

我有一张地图，里面有很多多边形，其中一个里面有一个点，如下所示： 多边形

多边形边缘的 x 和 y 坐标保存在这样的数据库中（例如）：

Polygon(Point(11824, 10756), Point(11822, 10618), Point(11912, 10517), Point(12060, 10529), Point(12158, 10604), Point(12133, 10713), Point(12027, 10812), Point(11902, 10902)),
Polygon(Point(11077, 13610), Point(10949, 13642), Point(10828, 13584), Point(10772, 13480), Point(10756, 13353), Point(10849, 13256), Point(10976, 13224), Point(11103, 13294), Point(11171, 13414), Point(11135, 13558)),
Polygon(Point(11051.801757813, 11373.985351563), Point(11165.717773438, 11275.469726563), Point(11281.733398438, 11255.646484375), Point(11381.07421875, 11333.15625), Point(11440.202148438, 11467.706054688), Point(11404.73046875, 11584.534179688), Point(11301.662109375, 11643.852539063), Point(11169.486328125, 11644.079101563), Point(11067.555664063, 11579.676757813), Point(11018.21484375, 11454.750976563)),
Polygon(Point(12145, 13013), Point(12069.065429688, 13014.67578125), Point(12012.672851563, 12953.833984375), Point(11973.942382813, 12910.14453125), Point(11958.610351563, 12853.736328125), Point(11988.58203125, 12780.668945313), Point(12046.806640625, 12735.046875), Point(12117.080078125, 12729.838867188), Point(12185.567382813, 12743.389648438), Point(12225.575195313, 12803.530273438), Point(12255.934570313, 12859.2109375), Point(12263.861328125, 12914.166992188), Point(12221.2578125, 12978.983398438)),

它们是稍后绘制的，我无法访问它，只能访问这个坐标。所以我有多边形边缘的 x/y 和红点的 x/y。现在我必须知道红点在哪个多边形中。

那么我需要的最重要的是： 多边形

我有红点的 x 和 y 坐标以及边缘的 xy 坐标。我需要绿点的 x 和 y 坐标。（多边形外最近的位置）

我在lua中需要它，但随时用任何语言回答，我可以翻译它。

4

3 回答 3

7

要知道该点在哪个多边形中，您可以使用Ray Casting algorithm。

为了找到最近的边缘，您可以使用一种简单的方法来计算到每个边缘的距离，然后取最小值。只要记住检查与边缘的交点是否在边缘之外（检查这个）。如果它在外面，则距离边缘最近的极端。

好的，用一些伪代码更好地解释直觉：

dot(u,v) --> ((u).x * (v).x + (u).y * (v).y)
norm(v)  --> sqrt(dot(v,v))     // norm = length of vector
dist(u,v)--> norm(u-v)          // distance = norm of difference

// Vector contains x and y
// Point contains x and y
// Segment contains P0 and P1 of type Point
// Point  = Point ± Vector
// Vector = Point - Point
// Vector = Scalar * Vector
Point closest_Point_in_Segment_to(Point P, Segment S)
{
     Vector v = S.P1 - S.P0;
     Vector w = P - S.P0;

     double c1 = dot(w,v);
     if ( c1 <= 0 )   // the closest point is outside the segment and nearer to P0
          return S.P0;

     double c2 = dot(v,v);
     if ( c2 <= c1 )  // the closest point is outside the segment and nearer to P1
          return S.P1;

     double b = c1 / c2;
     Point Pb = S.P0 + b * v;
     return Pb;
}

[Point, Segment] get_closest_border_point_to(Point point, Polygon poly) {

    double bestDistance = MAX_DOUBLE;
    Segment bestSegment;
    Point bestPoint;

    foreach s in poly.segments {
        Point closestInS = closest_Point_in_Segment_to(point, s);
        double d = dist(point, closestInS);
        if (d < bestDistance) {
            bestDistance = d;
            bestSegment = s;
            bestPoint = closestInS; 
        }
    }

    return [bestPoint, bestSegment];
}

我认为这个伪代码应该让你继续前进，当然，一旦你有了多边形，你的观点就在！

于 2013-09-27T10:16:23.063 回答
2

我的PolyCollisions班级：

public class PolyCollisions {

    // Call this function...
    public static Vector2 doCollisions (Vector2[] polygon, Vector2 point) {

        if(!pointIsInPoly(polygon, point)) {
            // The point is not colliding with the polygon, so it does not need to change location
            return point;
        }

        // Get the closest point of the polygon
        return closestPointOutsidePolygon(polygon, point);

    }

    // Check if the given point is within the given polygon (Vertexes)
    // 
    // If so, call on collision if required, and move the point to the
    // closest point outside of the polygon
    public static boolean pointIsInPoly(Vector2[] verts, Vector2 p) {
        int nvert = verts.length;
        double[] vertx = new double[nvert];
        double[] verty = new double[nvert];
        for(int i = 0; i < nvert; i++) {
            Vector2 vert = verts[i];
            vertx[i] = vert.x;
            verty[i] = vert.y;
        }
        double testx = p.x;
        double testy = p.y;
        int i, j;
        boolean c = false;
        for (i = 0, j = nvert-1; i < nvert; j = i++) {
            if ( ((verty[i]>testy) != (verty[j]>testy)) &&
                    (testx < (vertx[j]-vertx[i]) * (testy-verty[i]) / (verty[j]-verty[i]) + vertx[i]) )
                c = !c;
        }
        return c;
    }

    // Gets the closed point that isn't inside the polygon...
    public static Vector2 closestPointOutsidePolygon (Vector2[] poly, Vector2 point) {

        return getClosestPointInSegment(closestSegment(poly, point), point);

    }

    public static Vector2 getClosestPointInSegment (Vector2[] segment, Vector2 point) {

        return newPointFromCollision(segment[0], segment[1], point);

    }

    public static Vector2 newPointFromCollision (Vector2 aLine, Vector2 bLine, Vector2 p) {

        return nearestPointOnLine(aLine.x, aLine.y, bLine.x, bLine.y, p.x, p.y);

    }

    public static Vector2 nearestPointOnLine(double ax, double ay, double bx, double by, double px, double py) {

        // https://stackoverflow.com/questions/1459368/snap-point-to-a-line-java

        double apx = px - ax;
        double apy = py - ay;
        double abx = bx - ax;
        double aby = by - ay;

        double ab2 = abx * abx + aby * aby;
        double ap_ab = apx * abx + apy * aby;
        double t = ap_ab / ab2;
        if (t < 0) {
            t = 0;
        } else if (t > 1) {
            t = 1;
        }
        return new Vector2(ax + abx * t, ay + aby * t);
    }

    public static Vector2[] closestSegment (Vector2[] points, Vector2 point) {

        Vector2[] returns = new Vector2[2];

        int index = closestPointIndex(points, point);

        returns[0] = points[index];

        Vector2[] neighbors = new Vector2[] {
                points[(index+1+points.length)%points.length],
                points[(index-1+points.length)%points.length]
        };

        double[] neighborAngles = new double[] {
                getAngle(new Vector2[] {point, returns[0], neighbors[0]}),
                getAngle(new Vector2[] {point, returns[0], neighbors[1]})
        };
        // The neighbor with the lower angle is the one to use
        if(neighborAngles[0] < neighborAngles[1]) {
            returns[1] = neighbors[0];
        } else {
            returns[1] = neighbors[1];
        }

        return returns;

    }

    public static double getAngle (Vector2[] abc) {

        // https://stackoverflow.com/questions/1211212/how-to-calculate-an-angle-from-three-points
        // atan2(P2.y - P1.y, P2.x - P1.x) - atan2(P3.y - P1.y, P3.x - P1.x)
        return Math.atan2(abc[2].y - abc[0].y, abc[2].x - abc[0].x) - Math.atan2(abc[1].y - abc[0].y, abc[1].x - abc[0].x);

    }

    public static int closestPointIndex (Vector2[] points, Vector2 point) {

        int leastDistanceIndex = 0;
        double leastDistance = Double.MAX_VALUE;

        for(int i = 0; i < points.length; i++) {
            double dist = distance(points[i], point);
            if(dist < leastDistance) {
                leastDistanceIndex = i;
                leastDistance = dist;
            }
        }

        return leastDistanceIndex;

    }

    public static double distance (Vector2 a, Vector2 b) {
        return Math.sqrt(Math.pow(Math.abs(a.x-b.x), 2)+Math.pow(Math.abs(a.y-b.y), 2));
    }

}

这是一个小解释（有趣的事实：这是我发布到堆栈溢出的第一张图片！）

在此处输入图像描述

抱歉，太乱了……

课程分步说明：

  • 检查给定点是否在多边形内
    • 如果不是，则返回当前点，因为不需要更改。
  • 找到多边形最近的 VERTEX
    • 这不是最近的 POINT，因为一个点可以在顶点之间
  • 抓住顶点的两个邻居，保留一个角度较小的邻居。
    • 较低的角度比较高的角度具有更短的距离，因为较高的角度“消失”得更快
  • 使用StackOverflow上此问题的答案获取线段的最近点。

恭喜！你在糟糕的教程中幸存下来！希望它有所帮助:) +感谢所有帮助我帮助你的答案的评论链接！

将点捕捉到线

如何从3点计算角度

于 2017-07-15T23:38:50.907 回答
0

这是答案的 C# 版本@nathanfranke

using System;
using Windows.Foundation;
using Windows.UI.Xaml.Shapes;
using System.Numerics;

    public class PolyCollisions
        {
    
            // Call this function...
            public static Vector2 DoCollisions(Vector2[] polygon, Vector2 point)
            {
    
                if (!PointIsInPoly(polygon, point))
                {
                    // The point is not colliding with the polygon, so it does not need to change location
                    return point;
                }
    
                // Get the closest point of the polygon
                return ClosestPointOutsidePolygon(polygon, point);
    
            }
    
            // Check if the given point is within the given polygon (Vertexes)
            // 
            // If so, call on collision if required, and move the point to the
            // closest point outside of the polygon
            public static bool PointIsInPoly(Vector2[] verts, Vector2 p)
            {
                int nvert = verts.Length;
                double[] vertx = new double[nvert];
                double[] verty = new double[nvert];
                for (int d = 0; d < nvert; d++)
                {
                    Vector2 vert = verts[d];
                    vertx[d] = vert.X;
                    verty[d] = vert.Y;
                }
                
                double testx = p.X;
                double testy = p.Y;
                int i, j;
                bool c = false;
                for (i = 0, j = nvert - 1; i < nvert; j = i++)
                {
                    if (((verty[i] > testy) != (verty[j] > testy)) && (testx < (vertx[j] - vertx[i]) * (testy - verty[i]) / (verty[j] - verty[i]) + vertx[i]))
                        c = !c;
                }
                return c;
            }
    
            // Gets the closed point that isn't inside the polygon...
            public static Vector2 ClosestPointOutsidePolygon(Vector2[] poly, Vector2 point)
            {
                return GetClosestPointInSegment(ClosestSegment(poly, point), point);
            }
    
            public static Vector2 GetClosestPointInSegment(Vector2[] segment, Vector2 point)
            {
                return NewPointFromCollision(segment[0], segment[1], point);
            }
    
            public static Vector2 NewPointFromCollision(Vector2 aLine, Vector2 bLine, Vector2 p)
            {
                return NearestPointOnLine(aLine.X, aLine.Y, bLine.X, bLine.Y, p.X, p.Y);
            }
    
            public static Vector2 NearestPointOnLine(double ax, double ay, double bx, double by, double px, double py)
            {
                // https://stackoverflow.com/questions/1459368/snap-point-to-a-line-java
                double apx = px - ax;
                double apy = py - ay;
                double abx = bx - ax;
                double aby = by - ay;
    
                double ab2 = abx * abx + aby * aby;
                double ap_ab = apx * abx + apy * aby;
                double t = ap_ab / ab2;
                if (t < 0)
                {
                    t = 0;
                }
                else if (t > 1)
                {
                    t = 1;
                }
                return new Vector2((float)(ax + (abx * t)), (float)(ay + aby * t));
            }
    
            public static Vector2[] ClosestSegment(Vector2[] points, Vector2 point)
            {
    
                Vector2[] returns = new Vector2[2];
    
                int index = ClosestPointIndex(points, point);
    
                returns[0] = points[index];
    
                Vector2[] neighbors = new Vector2[] {
                    points[(index+1+points.Length)%points.Length],
                    points[(index-1+points.Length)%points.Length]
            };
    
                double[] neighborAngles = new double[] {
                    GetAngle(new Vector2[] {point, returns[0], neighbors[0]}),
                    GetAngle(new Vector2[] {point, returns[0], neighbors[1]})
            };
                // The neighbor with the lower angle is the one to use
                if (neighborAngles[0] < neighborAngles[1])
                {
                    returns[1] = neighbors[0];
                }
                else
                {
                    returns[1] = neighbors[1];
                }
    
                return returns;
    
            }
    
            public static double GetAngle(Vector2[] abc)
            {
    
                // https://stackoverflow.com/questions/1211212/how-to-calculate-an-angle-from-three-points
                // atan2(P2.y - P1.y, P2.x - P1.x) - atan2(P3.y - P1.y, P3.x - P1.x)
                return Math.Atan2(abc[2].X - abc[0].Y, abc[2].X - abc[0].X) - Math.Atan2(abc[1].Y - abc[0].Y, abc[1].X - abc[0].X);
    
            }
    
            public static int ClosestPointIndex(Vector2[] points, Vector2 point)
            {
    
                int leastDistanceIndex = 0;
                double leastDistance = Double.MaxValue;
    
                for (int i = 0; i < points.Length; i++)
                {
                    double dist = Distance(points[i], point);
                    if (dist < leastDistance)
                    {
                        leastDistanceIndex = i;
                        leastDistance = dist;
                    }
                }
    
                return leastDistanceIndex;
    
            }
    
            public static double Distance(Vector2 a, Vector2 b)
            {
                return Math.Sqrt(Math.Pow(Math.Abs(a.X - b.X), 2) + Math.Pow(Math.Abs(a.Y - b.Y), 2));
            }
    
        }
于 2021-06-22T01:15:32.820 回答