C#

改编自@Grumdrig

public static double MinimumDistanceToLineSegment(this Point p,
    Line line)
{
    var v = line.StartPoint;
    var w = line.EndPoint;

    double lengthSquared = DistanceSquared(v, w);

    if (lengthSquared == 0.0)
        return Distance(p, v);

    double t = Math.Max(0, Math.Min(1, DotProduct(p - v, w - v) / lengthSquared));
    var projection = v + t * (w - v);

    return Distance(p, projection);
}

public static double Distance(Point a, Point b)
{
    return Math.Sqrt(DistanceSquared(a, b));
}

public static double DistanceSquared(Point a, Point b)
{
    var d = a - b;
    return DotProduct(d, d);
}

public static double DotProduct(Point a, Point b)
{
    return (a.X * b.X) + (a.Y * b.Y);
}

只是遇到了这个，我想我应该添加一个Lua实现。它假设点以表{x=xVal, y=yVal}给出，直线或线段由包含两个点的表给出(见下面的例子):

function distance( P1, P2 )
    return math.sqrt((P1.x-P2.x)^2 + (P1.y-P2.y)^2)
end

-- Returns false if the point lies beyond the reaches of the segment
function distPointToSegment( line, P )
    if line[1].x == line[2].x and line[1].y == line[2].y then
        print("Error: Not a line!")
        return false
    end

    local d = distance( line[1], line[2] )

    local t = ((P.x - line[1].x)*(line[2].x - line[1].x) + (P.y - line[1].y)*(line[2].y - line[1].y))/(d^2)

    local projection = {}
    projection.x = line[1].x + t*(line[2].x-line[1].x)
    projection.y = line[1].y + t*(line[2].y-line[1].y)

    if t >= 0 and t <= 1 then   -- within line segment?
        return distance( projection, {x=P.x, y=P.y} )
    else
        return false
    end
end

-- Returns value even if point is further down the line (outside segment)
function distPointToLine( line, P )
    if line[1].x == line[2].x and line[1].y == line[2].y then
        print("Error: Not a line!")
        return false
    end

    local d = distance( line[1], line[2] )

    local t = ((P.x - line[1].x)*(line[2].x - line[1].x) + (P.y - line[1].y)*(line[2].y - line[1].y))/(d^2)

    local projection = {}
    projection.x = line[1].x + t*(line[2].x-line[1].x)
    projection.y = line[1].y + t*(line[2].y-line[1].y)

    return distance( projection, {x=P.x, y=P.y} )
end

使用示例:

local P1 = {x = 0, y = 0}
local P2 = {x = 10, y = 10}
local line = { P1, P2 }
local P3 = {x = 7, y = 15}
print(distPointToLine( line, P3 ))  -- prints 5.6568542494924
print(distPointToSegment( line, P3 )) -- prints false

在f#中，点c到a和b之间的线段的距离为:

let pointToLineSegmentDistance (a: Vector, b: Vector) (c: Vector) =
  let d = b - a
  let s = d.Length
  let lambda = (c - a) * d / s
  let p = (lambda |> max 0.0 |> min s) * d / s
  (a + p - c).Length

向量d沿着线段从a指向b。d/s与c-a的点积给出了无限直线与点c之间最接近点的参数。使用min和max函数将该参数钳制到范围0..s，使该点位于a和b之间。最后，a+p-c的长度是c到线段上最近点的距离。

使用示例:

pointToLineSegmentDistance (Vector(0.0, 0.0), Vector(1.0, 0.0)) (Vector(-1.0, 1.0))

现在我的解决方案...... (Javascript)

这是非常快的，因为我试图避免任何数学。战俘的功能。

如你所见，在函数的最后，我得到了直线的距离。

代码来自lib http://www.draw2d.org/graphiti/jsdoc/#!/例子

/**
 * Static util function to determine is a point(px,py) on the line(x1,y1,x2,y2)
 * A simple hit test.
 * 
 * @return {boolean}
 * @static
 * @private
 * @param {Number} coronaWidth the accepted corona for the hit test
 * @param {Number} X1 x coordinate of the start point of the line
 * @param {Number} Y1 y coordinate of the start point of the line
 * @param {Number} X2 x coordinate of the end point of the line
 * @param {Number} Y2 y coordinate of the end point of the line
 * @param {Number} px x coordinate of the point to test
 * @param {Number} py y coordinate of the point to test
 **/
graphiti.shape.basic.Line.hit= function( coronaWidth, X1, Y1,  X2,  Y2, px, py)
{
  // Adjust vectors relative to X1,Y1
  // X2,Y2 becomes relative vector from X1,Y1 to end of segment
  X2 -= X1;
  Y2 -= Y1;
  // px,py becomes relative vector from X1,Y1 to test point
  px -= X1;
  py -= Y1;
  var dotprod = px * X2 + py * Y2;
  var projlenSq;
  if (dotprod <= 0.0) {
      // px,py is on the side of X1,Y1 away from X2,Y2
      // distance to segment is length of px,py vector
      // "length of its (clipped) projection" is now 0.0
      projlenSq = 0.0;
  } else {
      // switch to backwards vectors relative to X2,Y2
      // X2,Y2 are already the negative of X1,Y1=>X2,Y2
      // to get px,py to be the negative of px,py=>X2,Y2
      // the dot product of two negated vectors is the same
      // as the dot product of the two normal vectors
      px = X2 - px;
      py = Y2 - py;
      dotprod = px * X2 + py * Y2;
      if (dotprod <= 0.0) {
          // px,py is on the side of X2,Y2 away from X1,Y1
          // distance to segment is length of (backwards) px,py vector
          // "length of its (clipped) projection" is now 0.0
          projlenSq = 0.0;
      } else {
          // px,py is between X1,Y1 and X2,Y2
          // dotprod is the length of the px,py vector
          // projected on the X2,Y2=>X1,Y1 vector times the
          // length of the X2,Y2=>X1,Y1 vector
          projlenSq = dotprod * dotprod / (X2 * X2 + Y2 * Y2);
      }
  }
    // Distance to line is now the length of the relative point
    // vector minus the length of its projection onto the line
    // (which is zero if the projection falls outside the range
    //  of the line segment).
    var lenSq = px * px + py * py - projlenSq;
    if (lenSq < 0) {
        lenSq = 0;
    }
    return Math.sqrt(lenSq)<coronaWidth;
};

在我自己的问题线程如何计算在C, c# / .NET 2.0或Java的所有情况下一个点和线段之间的最短2D距离?当我找到一个c#的答案时，我被要求把它放在这里:所以它是从http://www.topcoder.com/tc?d1=tutorials&d2=geometry1&module=Static修改的:

//Compute the dot product AB . BC
private double DotProduct(double[] pointA, double[] pointB, double[] pointC)
{
    double[] AB = new double[2];
    double[] BC = new double[2];
    AB[0] = pointB[0] - pointA[0];
    AB[1] = pointB[1] - pointA[1];
    BC[0] = pointC[0] - pointB[0];
    BC[1] = pointC[1] - pointB[1];
    double dot = AB[0] * BC[0] + AB[1] * BC[1];

    return dot;
}

//Compute the cross product AB x AC
private double CrossProduct(double[] pointA, double[] pointB, double[] pointC)
{
    double[] AB = new double[2];
    double[] AC = new double[2];
    AB[0] = pointB[0] - pointA[0];
    AB[1] = pointB[1] - pointA[1];
    AC[0] = pointC[0] - pointA[0];
    AC[1] = pointC[1] - pointA[1];
    double cross = AB[0] * AC[1] - AB[1] * AC[0];

    return cross;
}

//Compute the distance from A to B
double Distance(double[] pointA, double[] pointB)
{
    double d1 = pointA[0] - pointB[0];
    double d2 = pointA[1] - pointB[1];

    return Math.Sqrt(d1 * d1 + d2 * d2);
}

//Compute the distance from AB to C
//if isSegment is true, AB is a segment, not a line.
double LineToPointDistance2D(double[] pointA, double[] pointB, double[] pointC, 
    bool isSegment)
{
    double dist = CrossProduct(pointA, pointB, pointC) / Distance(pointA, pointB);
    if (isSegment)
    {
        double dot1 = DotProduct(pointA, pointB, pointC);
        if (dot1 > 0) 
            return Distance(pointB, pointC);

        double dot2 = DotProduct(pointB, pointA, pointC);
        if (dot2 > 0) 
            return Distance(pointA, pointC);
    }
    return Math.Abs(dist);
} 

我不是要回答问题，而是要问问题，所以我希望我不会因为某些原因而得到数百万张反对票，而是批评。我只是想(并被鼓励)分享其他人的想法，因为这个帖子中的解决方案要么是用一些奇异的语言(Fortran, Mathematica)，要么被某人标记为错误。对我来说唯一有用的(由Grumdrig编写)是用c++编写的，没有人标记它有错误。但是它缺少被调用的方法(dot等)。

用t-sql编码

点为(@px， @py)，线段从(@ax， @ay)到(@bx， @by)

create function fn_sqr (@NumberToSquare decimal(18,10)) 
returns decimal(18,10)
as 
begin
    declare @Result decimal(18,10)
    set @Result = @NumberToSquare * @NumberToSquare
    return @Result
end
go

create function fn_Distance(@ax decimal (18,10) , @ay decimal (18,10), @bx decimal(18,10),  @by decimal(18,10)) 
returns decimal(18,10)
as
begin
    declare @Result decimal(18,10)
    set @Result = (select dbo.fn_sqr(@ax - @bx) + dbo.fn_sqr(@ay - @by) )
    return @Result
end
go

create function fn_DistanceToSegmentSquared(@px decimal(18,10), @py decimal(18,10), @ax decimal(18,10), @ay decimal(18,10), @bx decimal(18,10), @by decimal(18,10)) 
returns decimal(18,10)
as 
begin
    declare @l2 decimal(18,10)
    set @l2 = (select dbo.fn_Distance(@ax, @ay, @bx, @by))
    if @l2 = 0
        return dbo.fn_Distance(@px, @py, @ax, @ay)
    declare @t decimal(18,10)
    set @t = ((@px - @ax) * (@bx - @ax) + (@py - @ay) * (@by - @ay)) / @l2
    if (@t < 0) 
        return dbo.fn_Distance(@px, @py, @ax, @ay);
    if (@t > 1) 
        return dbo.fn_Distance(@px, @py, @bx, @by);
    return dbo.fn_Distance(@px, @py,  @ax + @t * (@bx - @ax),  @ay + @t * (@by - @ay))
end
go

create function fn_DistanceToSegment(@px decimal(18,10), @py decimal(18,10), @ax decimal(18,10), @ay decimal(18,10), @bx decimal(18,10), @by decimal(18,10)) 
returns decimal(18,10)
as 
begin
    return sqrt(dbo.fn_DistanceToSegmentSquared(@px, @py , @ax , @ay , @bx , @by ))
end
go

--example execution for distance from a point at (6,1) to line segment that runs from (4,2) to (2,1)
select dbo.fn_DistanceToSegment(6, 1, 4, 2, 2, 1) 
--result = 2.2360679775

--example execution for distance from a point at (-3,-2) to line segment that runs from (0,-2) to (-2,1)
select dbo.fn_DistanceToSegment(-3, -2, 0, -2, -2, 1) 
--result = 2.4961508830

--example execution for distance from a point at (0,-2) to line segment that runs from (0,-2) to (-2,1)
select dbo.fn_DistanceToSegment(0,-2, 0, -2, -2, 1) 
--result = 0.0000000000