GLSL版:

// line (a -> b ) point p[enter image description here][1]
float distanceToLine(vec2 a, vec2 b, vec2 p) {
    float aside = dot((p - a),(b - a));
    if(aside< 0.0) return length(p-a);
    float bside = dot((p - b),(a - b));
    if(bside< 0.0) return length(p-b);
    vec2 pointOnLine = (bside*a + aside*b)/pow(length(a-b),2.0);
    return length(p - pointOnLine);
}

WPF版本:

public class LineSegment
{
    private readonly Vector _offset;
    private readonly Vector _vector;

    public LineSegment(Point start, Point end)
    {
        _offset = (Vector)start;
        _vector = (Vector)(end - _offset);
    }

    public double DistanceTo(Point pt)
    {
        var v = (Vector)pt - _offset;

        // first, find a projection point on the segment in parametric form (0..1)
        var p = (v * _vector) / _vector.LengthSquared;

        // and limit it so it lays inside the segment
        p = Math.Min(Math.Max(p, 0), 1);

        // now, find the distance from that point to our point
        return (_vector * p - v).Length;
    }
}

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|

伊莱，你选定的代码是错误的。在线段所在直线附近但远离线段一端的点将被错误地判断为接近线段。更新:上面提到的错误答案已不再被接受。

下面是一些正确的c++代码。它假设一个2d向量类vec2 {float x,y;}，本质上，带有加法、subract、缩放等运算符，以及一个距离和点积函数(即x1 x2 + y1 y2)。

float minimum_distance(vec2 v, vec2 w, vec2 p) {
  // Return minimum distance between line segment vw and point p
  const float l2 = length_squared(v, w);  // i.e. |w-v|^2 -  avoid a sqrt
  if (l2 == 0.0) return distance(p, v);   // v == w case
  // 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
  // We clamp t from [0,1] to handle points outside the segment vw.
  const float t = max(0, min(1, dot(p - v, w - v) / l2));
  const vec2 projection = v + t * (w - v);  // Projection falls on the segment
  return distance(p, projection);
}

编辑:我需要一个Javascript实现，所以在这里，没有依赖关系(或注释，但它是一个直接的端口以上)。点被表示为具有x和y属性的对象。

function sqr(x) { return x * x }
function dist2(v, w) { return sqr(v.x - w.x) + sqr(v.y - w.y) }
function distToSegmentSquared(p, v, w) {
  var l2 = dist2(v, w);
  if (l2 == 0) return dist2(p, v);
  var t = ((p.x - v.x) * (w.x - v.x) + (p.y - v.y) * (w.y - v.y)) / l2;
  t = Math.max(0, Math.min(1, t));
  return dist2(p, { x: v.x + t * (w.x - v.x),
                    y: v.y + t * (w.y - v.y) });
}
function distToSegment(p, v, w) { return Math.sqrt(distToSegmentSquared(p, v, w)); }

编辑2:我需要一个Java版本，但更重要的是，我需要3d版本而不是2d版本。

float dist_to_segment_squared(float px, float py, float pz, float lx1, float ly1, float lz1, float lx2, float ly2, float lz2) {
  float line_dist = dist_sq(lx1, ly1, lz1, lx2, ly2, lz2);
  if (line_dist == 0) return dist_sq(px, py, pz, lx1, ly1, lz1);
  float t = ((px - lx1) * (lx2 - lx1) + (py - ly1) * (ly2 - ly1) + (pz - lz1) * (lz2 - lz1)) / line_dist;
  t = constrain(t, 0, 1);
  return dist_sq(px, py, pz, lx1 + t * (lx2 - lx1), ly1 + t * (ly2 - ly1), lz1 + t * (lz2 - lz1));
}

这里，在函数参数中，<px,py,pz>是问题点，线段有端点<lx1,ly1,lz1>和<lx2,ly2,lz2>。函数dist_sq(假定存在)求两点之间距离的平方。

本想在GLSL中这样做，但如果可能的话，最好避免所有这些条件。使用clamp()可以避免两种端点情况:

// find closest point to P on line segment AB:
vec3 closest_point_on_line_segment(in vec3 P, in vec3 A, in vec3 B) {
    vec3 AP = P - A, AB = B - A;
    float l = dot(AB, AB);
    if (l <= 0.0000001) return A;    // A and B are practically the same
    return AP - AB*clamp(dot(AP, AB)/l, 0.0, 1.0);  // do the projection
}

如果您可以确定A和B彼此不会非常接近，则可以简化为删除If()。事实上，即使A和B是相同的，我的GPU仍然给出了这个无条件版本的正确结果(但这是使用pre-OpenGL 4.1，其中GLSL除零是未定义的):

// find closest point to P on line segment AB:
vec3 closest_point_on_line_segment(in vec3 P, in vec3 A, in vec3 B) {
    vec3 AP = P - A, AB = B - A;
    return AP - AB*clamp(dot(AP, AB)/dot(AB, AB), 0.0, 1.0);
}

计算距离是很简单的——GLSL提供了一个distance()函数，你可以在这个最近的点和P。

灵感来自Iñigo Quilez的胶囊距离函数代码

I'm assuming you want to find the shortest distance between the point and a line segment; to do this, you need to find the line (lineA) which is perpendicular to your line segment (lineB) which goes through your point, determine the intersection between that line (lineA) and your line which goes through your line segment (lineB); if that point is between the two points of your line segment, then the distance is the distance between your point and the point you just found which is the intersection of lineA and lineB; if the point is not between the two points of your line segment, you need to get the distance between your point and the closer of two ends of the line segment; this can be done easily by taking the square distance (to avoid a square root) between the point and the two points of the line segment; whichever is closer, take the square root of that one.