c*t
2 楼
easy. There is a scan line method.
Basically, let a horizontal line (or any arbitrary line, but horizontal
is easier computationally) go through the point. Then count the # of
interceptions of the curves on this line. If the point is inside the
curves, then the # of interceptions to the left of the point should be
odd.
，
【在 x*********l 的大作中提到】
: 请教大家一个算法的问题。在一个图像里有很多互相不相交的封闭曲线（任意形状），
: 如何判断任意一个点是在这些封闭曲线内，还是在曲线外？
Basically, let a horizontal line (or any arbitrary line, but horizontal
is easier computationally) go through the point. Then count the # of
interceptions of the curves on this line. If the point is inside the
curves, then the # of interceptions to the left of the point should be
odd.
，
【在 x*********l 的大作中提到】
: 请教大家一个算法的问题。在一个图像里有很多互相不相交的封闭曲线（任意形状），
: 如何判断任意一个点是在这些封闭曲线内，还是在曲线外？
g*g
3 楼
任意画条线，数交点单双。
，
【在 x*********l 的大作中提到】
: 请教大家一个算法的问题。在一个图像里有很多互相不相交的封闭曲线（任意形状），
: 如何判断任意一个点是在这些封闭曲线内，还是在曲线外？
，
【在 x*********l 的大作中提到】
: 请教大家一个算法的问题。在一个图像里有很多互相不相交的封闭曲线（任意形状），
: 如何判断任意一个点是在这些封闭曲线内，还是在曲线外？
x*l
4 楼
3x
【在 c*****t 的大作中提到】
: easy. There is a scan line method.
: Basically, let a horizontal line (or any arbitrary line, but horizontal
: is easier computationally) go through the point. Then count the # of
: interceptions of the curves on this line. If the point is inside the
: curves, then the # of interceptions to the left of the point should be
: odd.
:
: ，
【在 c*****t 的大作中提到】
: easy. There is a scan line method.
: Basically, let a horizontal line (or any arbitrary line, but horizontal
: is easier computationally) go through the point. Then count the # of
: interceptions of the curves on this line. If the point is inside the
: curves, then the # of interceptions to the left of the point should be
: odd.
:
: ，
l*u
5 楼
Sounds good. The only thing is that one needs to be careful in determining
if the line is intersecting, not just tangent, with the curve.
【在 c*****t 的大作中提到】
: easy. There is a scan line method.
: Basically, let a horizontal line (or any arbitrary line, but horizontal
: is easier computationally) go through the point. Then count the # of
: interceptions of the curves on this line. If the point is inside the
: curves, then the # of interceptions to the left of the point should be
: odd.
:
: ，
if the line is intersecting, not just tangent, with the curve.
【在 c*****t 的大作中提到】
: easy. There is a scan line method.
: Basically, let a horizontal line (or any arbitrary line, but horizontal
: is easier computationally) go through the point. Then count the # of
: interceptions of the curves on this line. If the point is inside the
: curves, then the # of interceptions to the left of the point should be
: odd.
:
: ，
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