我需要一个基本函数来求点到线段的最短距离。你可以随意用任何你想要的语言来编写解决方案;我可以把它翻译成什么我正在使用(Javascript)。
编辑:线段由两个端点定义。线段AB由两点A (x1,y1)和B (x2,y2)定义。我要求的是这条线段到点C (x3,y3)的距离。我的几何技能生疏了,所以我看到的例子让我很困惑,我很遗憾地承认。
我需要一个基本函数来求点到线段的最短距离。你可以随意用任何你想要的语言来编写解决方案;我可以把它翻译成什么我正在使用(Javascript)。
编辑:线段由两个端点定义。线段AB由两点A (x1,y1)和B (x2,y2)定义。我要求的是这条线段到点C (x3,y3)的距离。我的几何技能生疏了,所以我看到的例子让我很困惑,我很遗憾地承认。
当前回答
JavaScript中一个基于这个公式的更简洁的解决方案:
distToSegment: function (point, linePointA, linePointB){
var x0 = point.X;
var y0 = point.Y;
var x1 = linePointA.X;
var y1 = linePointA.Y;
var x2 = linePointB.X;
var y2 = linePointB.Y;
var Dx = (x2 - x1);
var Dy = (y2 - y1);
var numerator = Math.abs(Dy*x0 - Dx*y0 - x1*y2 + x2*y1);
var denominator = Math.sqrt(Dx*Dx + Dy*Dy);
if (denominator == 0) {
return this.dist2(point, linePointA);
}
return numerator/denominator;
}
其他回答
这是一个自成体系的Delphi / Pascal版本的函数,基于上面约书亚的答案。使用TPoint用于VCL屏幕图形,但应该易于根据需要进行调整。
function DistancePtToSegment( pt, pt1, pt2: TPoint): double;
var
a, b, c, d: double;
len_sq: double;
param: double;
xx, yy: double;
dx, dy: double;
begin
a := pt.x - pt1.x;
b := pt.y - pt1.y;
c := pt2.x - pt1.x;
d := pt2.y - pt1.y;
len_sq := (c * c) + (d * d);
param := -1;
if (len_sq <> 0) then
begin
param := ((a * c) + (b * d)) / len_sq;
end;
if param < 0 then
begin
xx := pt1.x;
yy := pt1.y;
end
else if param > 1 then
begin
xx := pt2.x;
yy := pt2.y;
end
else begin
xx := pt1.x + param * c;
yy := pt1.y + param * d;
end;
dx := pt.x - xx;
dy := pt.y - yy;
result := sqrt( (dx * dx) + (dy * dy))
end;
这是我最后写的代码。这段代码假设一个点以{x:5, y:7}的形式定义。注意,这不是绝对最有效的方法,但它是我能想到的最简单、最容易理解的代码。
// a, b, and c in the code below are all points
function distance(a, b)
{
var dx = a.x - b.x;
var dy = a.y - b.y;
return Math.sqrt(dx*dx + dy*dy);
}
function Segment(a, b)
{
var ab = {
x: b.x - a.x,
y: b.y - a.y
};
var length = distance(a, b);
function cross(c) {
return ab.x * (c.y-a.y) - ab.y * (c.x-a.x);
};
this.distanceFrom = function(c) {
return Math.min(distance(a,c),
distance(b,c),
Math.abs(cross(c) / length));
};
}
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);
}
这里没有看到Java实现,所以我将Javascript函数从接受的答案转换为Java代码:
static double sqr(double x) {
return x * x;
}
static double dist2(DoublePoint v, DoublePoint w) {
return sqr(v.x - w.x) + sqr(v.y - w.y);
}
static double distToSegmentSquared(DoublePoint p, DoublePoint v, DoublePoint w) {
double l2 = dist2(v, w);
if (l2 == 0) return dist2(p, v);
double t = ((p.x - v.x) * (w.x - v.x) + (p.y - v.y) * (w.y - v.y)) / l2;
if (t < 0) return dist2(p, v);
if (t > 1) return dist2(p, w);
return dist2(p, new DoublePoint(
v.x + t * (w.x - v.x),
v.y + t * (w.y - v.y)
));
}
static double distToSegment(DoublePoint p, DoublePoint v, DoublePoint w) {
return Math.sqrt(distToSegmentSquared(p, v, w));
}
static class DoublePoint {
public double x;
public double y;
public DoublePoint(double x, double y) {
this.x = x;
this.y = y;
}
}
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)
}