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
忍不住用python来编码:)
from math import sqrt, fabs
def pdis(a, b, c):
t = b[0]-a[0], b[1]-a[1] # Vector ab
dd = sqrt(t[0]**2+t[1]**2) # Length of ab
t = t[0]/dd, t[1]/dd # unit vector of ab
n = -t[1], t[0] # normal unit vector to ab
ac = c[0]-a[0], c[1]-a[1] # vector ac
return fabs(ac[0]*n[0]+ac[1]*n[1]) # Projection of ac to n (the minimum distance)
print pdis((1,1), (2,2), (2,0)) # Example (answer is 1.414)
fortran也是一样:)
real function pdis(a, b, c)
real, dimension(0:1), intent(in) :: a, b, c
real, dimension(0:1) :: t, n, ac
real :: dd
t = b - a ! Vector ab
dd = sqrt(t(0)**2+t(1)**2) ! Length of ab
t = t/dd ! unit vector of ab
n = (/-t(1), t(0)/) ! normal unit vector to ab
ac = c - a ! vector ac
pdis = abs(ac(0)*n(0)+ac(1)*n(1)) ! Projection of ac to n (the minimum distance)
end function pdis
program test
print *, pdis((/1.0,1.0/), (/2.0,2.0/), (/2.0,0.0/)) ! Example (answer is 1.414)
end program test
其他回答
嘿,我昨天才写的。它在Actionscript 3.0中,基本上是Javascript,尽管你可能没有相同的Point类。
//st = start of line segment
//b = the line segment (as in: st + b = end of line segment)
//pt = point to test
//Returns distance from point to line segment.
//Note: nearest point on the segment to the test point is right there if we ever need it
public static function linePointDist( st:Point, b:Point, pt:Point ):Number
{
var nearestPt:Point; //closest point on seqment to pt
var keyDot:Number = dot( b, pt.subtract( st ) ); //key dot product
var bLenSq:Number = dot( b, b ); //Segment length squared
if( keyDot <= 0 ) //pt is "behind" st, use st
{
nearestPt = st
}
else if( keyDot >= bLenSq ) //pt is "past" end of segment, use end (notice we are saving twin sqrts here cuz)
{
nearestPt = st.add(b);
}
else //pt is inside segment, reuse keyDot and bLenSq to get percent of seqment to move in to find closest point
{
var keyDotToPctOfB:Number = keyDot/bLenSq; //REM dot product comes squared
var partOfB:Point = new Point( b.x * keyDotToPctOfB, b.y * keyDotToPctOfB );
nearestPt = st.add(partOfB);
}
var dist:Number = (pt.subtract(nearestPt)).length;
return dist;
}
此外,这里有一个关于这个问题的相当完整和可读的讨论:notejot.com
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 );
}
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;
%Matlab solution by Tim from Cody
function ans=distP2S(x0,y0,x1,y1,x2,y2)
% Point is x0,y0
z=complex(x0-x1,y0-y1);
complex(x2-x1,y2-y1);
abs(z-ans*min(1,max(0,real(z/ans))));
Lua: 查找线段(不是整条线)与点之间的最小距离
function solveLinearEquation(A1,B1,C1,A2,B2,C2)
--it is the implitaion of a method of solving linear equations in x and y
local f1 = B1*C2 -B2*C1
local f2 = A2*C1-A1*C2
local f3 = A1*B2 -A2*B1
return {x= f1/f3, y= f2/f3}
end
function pointLiesOnLine(x,y,x1,y1,x2,y2)
local dx1 = x-x1
local dy1 = y-y1
local dx2 = x-x2
local dy2 = y-y2
local crossProduct = dy1*dx2 -dx1*dy2
if crossProduct ~= 0 then return false
else
if ((x1>=x) and (x>=x2)) or ((x2>=x) and (x>=x1)) then
if ((y1>=y) and (y>=y2)) or ((y2>=y) and (y>=y1)) then
return true
else return false end
else return false end
end
end
function dist(x1,y1,x2,y2)
local dx = x1-x2
local dy = y1-y2
return math.sqrt(dx*dx + dy* dy)
end
function findMinDistBetnPointAndLine(x1,y1,x2,y2,x3,y3)
-- finds the min distance between (x3,y3) and line (x1,y2)--(x2,y2)
local A2,B2,C2,A1,B1,C1
local dx = y2-y1
local dy = x2-x1
if dx == 0 then A2=1 B2=0 C2=-x3 A1=0 B1=1 C1=-y1
elseif dy == 0 then A2=0 B2=1 C2=-y3 A1=1 B1=0 C1=-x1
else
local m1 = dy/dx
local m2 = -1/m1
A2=m2 B2=-1 C2=y3-m2*x3 A1=m1 B1=-1 C1=y1-m1*x1
end
local intsecPoint= solveLinearEquation(A1,B1,C1,A2,B2,C2)
if pointLiesOnLine(intsecPoint.x, intsecPoint.y,x1,y1,x2,y2) then
return dist(intsecPoint.x, intsecPoint.y, x3,y3)
else
return math.min(dist(x3,y3,x1,y1),dist(x3,y3,x2,y2))
end
end