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
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;
其他回答
Lua解决方案
-- distance from point (px, py) to line segment (x1, y1, x2, y2)
function distPointToLine(px,py,x1,y1,x2,y2) -- point, start and end of the segment
local dx,dy = x2-x1,y2-y1
local length = math.sqrt(dx*dx+dy*dy)
dx,dy = dx/length,dy/length -- normalization
local p = dx*(px-x1)+dy*(py-y1)
if p < 0 then
dx,dy = px-x1,py-y1
return math.sqrt(dx*dx+dy*dy), x1, y1 -- distance, nearest point
elseif p > length then
dx,dy = px-x2,py-y2
return math.sqrt(dx*dx+dy*dy), x2, y2 -- distance, nearest point
end
return math.abs(dy*(px-x1)-dx*(py-y1)), x1+dx*p, y1+dy*p -- distance, nearest point
end
对于折线(有两条以上线段的线):
-- if the (poly-)line has several segments, just iterate through all of them:
function nearest_sector_in_line (x, y, line)
local x1, y1, x2, y2, min_dist
local ax,ay = line[1], line[2]
for j = 3, #line-1, 2 do
local bx,by = line[j], line[j+1]
local dist = distPointToLine(x,y,ax,ay,bx,by)
if not min_dist or dist < min_dist then
min_dist = dist
x1, y1, x2, y2 = ax,ay,bx,by
end
ax, ay = bx, by
end
return x1, y1, x2, y2
end
例子:
-- call it:
local x1, y1, x2, y2 = nearest_sector_in_line (7, 4, {0,0, 10,0, 10,10, 0,10})
下面是HSQLDB的SQL实现:
CREATE FUNCTION dist_to_segment(px double, py double, vx double, vy double, wx double, wy double)
RETURNS double
BEGIN atomic
declare l2 double;
declare t double;
declare nx double;
declare ny double;
set l2 =(vx - wx)*(vx - wx) + (vy - wy)*(vy - wy);
IF l2 = 0 THEN
RETURN sqrt((vx - px)*(vx - px) + (vy - py)*(vy - py));
ELSE
set t = ((px - vx) * (wx - vx) + (py - vy) * (wy - vy)) / l2;
set t = GREATEST(0, LEAST(1, t));
set nx=vx + t * (wx - vx);
set ny=vy + t * (wy - vy);
RETURN sqrt((nx - px)*(nx - px) + (ny - py)*(ny - py));
END IF;
END;
Postgres的实现:
CREATE FUNCTION dist_to_segment(px numeric, py numeric, vx numeric, vy numeric, wx numeric, wy numeric)
RETURNS numeric
AS $$
declare l2 numeric;
declare t numeric;
declare nx numeric;
declare ny numeric;
BEGIN
l2 := (vx - wx)*(vx - wx) + (vy - wy)*(vy - wy);
IF l2 = 0 THEN
RETURN sqrt((vx - px)*(vx - px) + (vy - py)*(vy - py));
ELSE
t := ((px - vx) * (wx - vx) + (py - vy) * (wy - vy)) / l2;
t := GREATEST(0, LEAST(1, t));
nx := vx + t * (wx - vx);
ny := vy + t * (wy - vy);
RETURN sqrt((nx - px)*(nx - px) + (ny - py)*(ny - py));
END IF;
END;
$$ LANGUAGE plpgsql;
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)
}
用Matlab直接实现Grumdrig
function ans=distP2S(px,py,vx,vy,wx,wy)
% [px py vx vy wx wy]
t=( (px-vx)*(wx-vx)+(py-vy)*(wy-vy) )/idist(vx,wx,vy,wy)^2;
[idist(px,vx,py,vy) idist(px,vx+t*(wx-vx),py,vy+t*(wy-vy)) idist(px,wx,py,wy) ];
ans(1+(t>0)+(t>1)); % <0 0<=t<=1 t>1
end
function d=idist(a,b,c,d)
d=abs(a-b+1i*(c-d));
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));
};
}