基于Joshua Javascript的AutoHotkeys版本:

plDist(x, y, x1, y1, x2, y2) {
    A:= x - x1
    B:= y - y1
    C:= x2 - x1
    D:= y2 - y1

    dot:= A*C + B*D
    sqLen:= C*C + D*D
    param:= dot / sqLen

    if (param < 0 || ((x1 = x2) && (y1 = y2))) {
        xx:= x1
        yy:= y1
    } else if (param > 1) {
        xx:= x2
        yy:= y2
    } else {
        xx:= x1 + param*C
        yy:= y1 + param*D
    }

    dx:= x - xx
    dy:= y - yy

    return sqrt(dx*dx + dy*dy)
}

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)。

以下是Grumdrig解决方案的一个更完整的说明。这个版本还返回最近的点本身。

#include "stdio.h"
#include "math.h"

class Vec2
{
public:
    float _x;
    float _y;

    Vec2()
    {
        _x = 0;
        _y = 0;
    }

    Vec2( const float x, const float y )
    {
        _x = x;
        _y = y;
    }

    Vec2 operator+( const Vec2 &v ) const
    {
        return Vec2( this->_x + v._x, this->_y + v._y );
    }

    Vec2 operator-( const Vec2 &v ) const
    {
        return Vec2( this->_x - v._x, this->_y - v._y );
    }

    Vec2 operator*( const float f ) const
    {
        return Vec2( this->_x * f, this->_y * f );
    }

    float DistanceToSquared( const Vec2 p ) const
    {
        const float dX = p._x - this->_x;
        const float dY = p._y - this->_y;

        return dX * dX + dY * dY;
    }

    float DistanceTo( const Vec2 p ) const
    {
        return sqrt( this->DistanceToSquared( p ) );
    }

    float DotProduct( const Vec2 p ) const
    {
        return this->_x * p._x + this->_y * p._y;
    }
};

// return minimum distance between line segment vw and point p, and the closest point on the line segment, q
float DistanceFromLineSegmentToPoint( const Vec2 v, const Vec2 w, const Vec2 p, Vec2 * const q )
{
    const float distSq = v.DistanceToSquared( w ); // i.e. |w-v|^2 ... avoid a sqrt
    if ( distSq == 0.0 )
    {
        // v == w case
        (*q) = v;

        return v.DistanceTo( p );
    }

    // consider the line extending the segment, parameterized as v + t (w - v)
    // we find projection of point p onto the line
    // it falls where t = [(p-v) . (w-v)] / |w-v|^2

    const float t = ( p - v ).DotProduct( w - v ) / distSq;
    if ( t < 0.0 )
    {
        // beyond the v end of the segment
        (*q) = v;

        return v.DistanceTo( p );
    }
    else if ( t > 1.0 )
    {
        // beyond the w end of the segment
        (*q) = w;

        return w.DistanceTo( p );
    }

    // projection falls on the segment
    const Vec2 projection = v + ( ( w - v ) * t );

    (*q) = projection;

    return p.DistanceTo( projection );
}

float DistanceFromLineSegmentToPoint( float segmentX1, float segmentY1, float segmentX2, float segmentY2, float pX, float pY, float *qX, float *qY )
{
    Vec2 q;

    float distance = DistanceFromLineSegmentToPoint( Vec2( segmentX1, segmentY1 ), Vec2( segmentX2, segmentY2 ), Vec2( pX, pY ), &q );

    (*qX) = q._x;
    (*qY) = q._y;

    return distance;
}

void TestDistanceFromLineSegmentToPoint( float segmentX1, float segmentY1, float segmentX2, float segmentY2, float pX, float pY )
{
    float qX;
    float qY;
    float d = DistanceFromLineSegmentToPoint( segmentX1, segmentY1, segmentX2, segmentY2, pX, pY, &qX, &qY );
    printf( "line segment = ( ( %f, %f ), ( %f, %f ) ), p = ( %f, %f ), distance = %f, q = ( %f, %f )\n",
            segmentX1, segmentY1, segmentX2, segmentY2, pX, pY, d, qX, qY );
}

void TestDistanceFromLineSegmentToPoint()
{
    TestDistanceFromLineSegmentToPoint( 0, 0, 1, 1, 1, 0 );
    TestDistanceFromLineSegmentToPoint( 0, 0, 20, 10, 5, 4 );
    TestDistanceFromLineSegmentToPoint( 0, 0, 20, 10, 30, 15 );
    TestDistanceFromLineSegmentToPoint( 0, 0, 20, 10, -30, 15 );
    TestDistanceFromLineSegmentToPoint( 0, 0, 10, 0, 5, 1 );
    TestDistanceFromLineSegmentToPoint( 0, 0, 0, 10, 1, 5 );
}

一个2D和3D的解决方案

考虑基底的变化，使得线段变成(0,0,0)-(d, 0,0)和点(u, v, 0)。在这个平面上，最短的距离由

    u ≤ 0 -> d(A, C)
0 ≤ u ≤ d -> |v|
d ≤ u     -> d(B, C)

(到其中一个端点或到支撑线的距离，取决于到该线的投影。等距轨迹由两个半圆和两条线段组成。)

式中，d为AB线段的长度，u、v分别为AB/d (AB方向的单位矢量)与AC的标量积和外积的模量。

AB.AC ≤ 0             -> |AC|
    0 ≤ AB.AC ≤ AB²   -> |ABxAC|/|AB|
          AB² ≤ AB.AC -> |BC|

请参见以下网站中的Matlab几何工具箱: http://people.sc.fsu.edu/~jburkardt/m_src/geometry/geometry.html

按Ctrl +f，输入“segment”，查找线段相关函数。函数“segment_point_dist_2d.”和segment_point_dist_3d。M "是你需要的。

几何代码有C版本、c++版本、FORTRAN77版本、FORTRAN90版本和MATLAB版本。

这里是与c++答案相同的东西，但移植到pascal。点参数的顺序已经改变，以适应我的代码，但还是一样的东西。

function Dot(const p1, p2: PointF): double;
begin
  Result := p1.x * p2.x + p1.y * p2.y;
end;
function SubPoint(const p1, p2: PointF): PointF;
begin
  result.x := p1.x - p2.x;
  result.y := p1.y - p2.y;
end;

function ShortestDistance2(const p,v,w : PointF) : double;
var
  l2,t : double;
  projection,tt: PointF;
begin
  // Return minimum distance between line segment vw and point p
  //l2 := length_squared(v, w);  // i.e. |w-v|^2 -  avoid a sqrt
  l2 := Distance(v,w);
  l2 := MPower(l2,2);
  if (l2 = 0.0) then begin
    result:= Distance(p, v);   // v == w case
    exit;
  end;
  // Consider the line extending the segment, parameterized as v + t (w - v).
  // We find projection of point p onto the line.
  // It falls where t = [(p-v) . (w-v)] / |w-v|^2
  t := Dot(SubPoint(p,v),SubPoint(w,v)) / l2;
  if (t < 0.0) then begin
    result := Distance(p, v);       // Beyond the 'v' end of the segment
    exit;
  end
  else if (t > 1.0) then begin
    result := Distance(p, w);  // Beyond the 'w' end of the segment
    exit;
  end;
  //projection := v + t * (w - v);  // Projection falls on the segment
  tt.x := v.x + t * (w.x - v.x);
  tt.y := v.y + t * (w.y - v.y);
  result := Distance(p, tt);
end;