如何从多边形内的点获取多边形外的最近点？

Question

我有一个包含很多多边形的地图，其中一个有一个点，如下所示：

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)),

它们是后来绘制的，我无法访问它，只能访问此坐标。 所以我有多边形边的x / y和红点的x / y。 现在我必须知道红点是哪个多边形。

那么我需要的最重要的是：

我有红点的x和y坐标以及边的xy坐标。 我需要绿点的x和y坐标。（多边形外面的最近位置）

我需要它在lua，但随意用任何语言回答，我可以翻译它。

Answer 1

要知道点在哪个多边形中，可以使用Ray Casting算法 。

为了找到最近的边缘，您可以使用简单的方法计算到每条边的距离 ，然后采用最小值。 只记得检查边缘的交点是否在边缘之外（检查一下 ）。 如果它在外面，则距离边缘最近的极端。

好吧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去，当然，一旦你有了多边形你的观点！

Answer 2

我的PolyCollisions课程：

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));
    }

}

这是一个小的解释（有趣的事实：这是我发布到堆栈溢出的第一个图像！）

对不起，它太乱了......

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课程一步一步：

• 检查给定点是否在多边形内
  • 如果不是，则返回当前点，因为不需要进行任何更改。
• 找到最接近多边形的VERTEX
  • 这不是最接近的POINT，因为点可以在顶点之间
• 抓住顶点的两个邻居，保持一个较低的角度。
  • 较低的角度具有比较高的角度更小的距离，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。

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恭喜！ 你幸免于糟糕的教程！ 希望它有所帮助:) +感谢所有评论链接到答案，帮助我帮助你！

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