如何在一条曲线上，获取到距离指定点最近的点位置？
与上一篇 C# 曲线上的点（一） 获取指定横坐标对应的纵坐标值 类似，
我们通过曲线上获取的密集点，通过俩点之间连线，获取连线上最近的点。我们能够获取到一系列最近的点集，最近只取距离最小的点即可。
我们这样的算法是否精确呢？不算太精确，但是对于获取曲线上最近点，基本能满足。
斜率变化不大的线段，点不密集；斜率变化较大的线段，点相当密集，所以由此点集得到的最近点，是相对准确的。
实现方案，以下代码可以直接复用：
1 public static Point GetClosestPointOnPath(Point p, Geometry geometry) 2 { 3 PathGeometry pathGeometry = geometry.GetFlattenedPathGeometry(); 4 5 var points = pathGeometry.Figures.Select(f => GetClosestPointOnPathFigure(f, p)) 6 .OrderBy(t => t.Item2).FirstOrDefault(); 7 return points?.Item1 ?? new Point(0, 0); 8 } 9 10 private static Tuple<Point, double> GetClosestPointOnPathFigure(PathFigure figure, Point p) 11 { 12 List<Tuple<Point, double>> closePoints = new List<Tuple<Point, double>>(); 13 Point current = figure.StartPoint; 14 foreach (PathSegment s in figure.Segments) 15 { 16 PolyLineSegment segment = s as PolyLineSegment; 17 LineSegment line = s as LineSegment; 18 Point[] points; 19 if (segment != null) 20 { 21 points = segment.Points.ToArray(); 22 } 23 else if (line != null) 24 { 25 points = new[] { line.Point }; 26 } 27 else 28 { 29 throw new InvalidOperationException(); 30 } 31 foreach (Point next in points) 32 { 33 Point closestPoint = GetClosestPointOnLine(current, next, p); 34 double d = (closestPoint - p).LengthSquared; 35 closePoints.Add(new Tuple<Point, double>(closestPoint, d)); 36 current = next; 37 } 38 } 39 return closePoints.OrderBy(t => t.Item2).First(); 40 } 41 42 private static Point GetClosestPointOnLine(Point start, Point end, Point p) 43 { 44 double length = (start - end).LengthSquared; 45 if (Math.Abs(length) < 0.01) 46 { 47 return start; 48 } 49 Vector v = end - start; 50 double param = (p - start) * v / length; 51 return (param < 0.0) ? start : (param > 1.0) ? end : (start + param * v); 52 }