获得离线最近的点

问题描述

我想要一个简单的 C# 函数来获取最近的点(从点 P)到线段 AB.一个抽象函数可能看起来像这样.我搜索了 SO，但没有找到可用的(由我自己)解决方案.

I'd like to have a straight forward C# function to get a closest point (from a point P) to a line-segment, AB. An abstract function may look like this. I've search through SO but not found a usable (by me) solution.

public Point getClosestPointFromLine(Point A, Point B, Point P);

推荐答案

这里是伪装成伪代码的 Ruby，假设 Point 对象每个都有一个 x 和 y 字段.

Here's Ruby disguised as Pseudo-Code, assuming Point objects each have a x and y field.

def GetClosestPoint(A, B, P)

  a_to_p = [P.x - A.x, P.y - A.y]     # Storing vector A->P
  a_to_b = [B.x - A.x, B.y - A.y]     # Storing vector A->B

  atb2 = a_to_b[0]**2 + a_to_b[1]**2  # **2 means "squared"
                                      #   Basically finding the squared magnitude
                                      #   of a_to_b

  atp_dot_atb = a_to_p[0]*a_to_b[0] + a_to_p[1]*a_to_b[1]
                                      # The dot product of a_to_p and a_to_b

  t = atp_dot_atb / atb2              # The normalized "distance" from a to
                                      #   your closest point

  return Point.new( :x => A.x + a_to_b[0]*t,
                    :y => A.y + a_to_b[1]*t )
                                      # Add the distance to A, moving
                                      #   towards B

end

<小时>

或者:

来自 Line-Line Intersection，维基百科.首先，找到 Q，这是从 P 朝着正确方向"迈出的第二个点.这给了我们四点.

From Line-Line Intersection, at Wikipedia. First, find Q, which is a second point that is to be had from taking a step from P in the "right direction". This gives us four points.

def getClosestPointFromLine(A, B, P)

  a_to_b = [B.x - A.x, B.y - A.y]   # Finding the vector from A to B
                                        This step can be combined with the next
  perpendicular = [ -a_to_b[1], a_to_b[0] ]
                                    # The vector perpendicular to a_to_b;
                                        This step can also be combined with the next

  Q = Point.new(:x => P.x + perpendicular[0], :y => P.y + perpendicular[1])
                                    # Finding Q, the point "in the right direction"
                                    # If you want a mess, you can also combine this
                                    # with the next step.

  return Point.new (:x => ((A.x*B.y - A.y*B.x)*(P.x - Q.x) - (A.x-B.x)*(P.x*Q.y - P.y*Q.x)) / ((A.x - B.x)*(P.y-Q.y) - (A.y - B.y)*(P.y-Q.y)),
                    :y => ((A.x*B.y - A.y*B.x)*(P.y - Q.y) - (A.y-B.y)*(P.x*Q.y - P.y*Q.x)) / ((A.x - B.x)*(P.y-Q.y) - (A.y - B.y)*(P.y-Q.y)) )

end

出于性能原因，缓存、跳过步骤等是可能的.

Caching, Skipping steps, etc. is possible, for performance reasons.

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