有没有办法在四叶分区找到相邻的细胞?
If I have an algorithm which is comprised of (let s say) three sub-algorithms, all with different O() characteristics, e.g.: algorithm A: O(n) algorithm B: O(log(n)) algorithm C: O(n log(n)) How do ...
有没有办法在四叶分区找到相邻的细胞?
空间填充曲线填满了空间填充, 并将2维度减为1维。 我在phpclasses. org (hilbert curve) 写了免费的php类。 它包括一个 z 曲线、 4 Hhilbert curps、 4 moore 曲线和一个四重函数 。 以下是关于碰撞探测和四重树的博客: Lab.polygonal.de/? p= 202?
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momonton a.k.a.z-curve 很容易构建。 将 x 和 y 值转换为二进制和组合值。 您可以在这里找到一些代码 。 您可以在此找到一些代码 : < a href="http://msdn. microsoft. com/ en- us/library/bb259689. aspx" rel= “ nofollow noreferrer" > > http://msdn. microsoft. com/ en-us/library/b259689. aspx < /a > 。 您可以使用最重要的部分来验证上方边界 。
您需要跟踪哪个孩子的节点。 如果相邻的节点位于同一个父系, 请返回它。 否则, 您需要在树上行走, 直到找到共同的祖先。 然后沿着类似的路径向下走, 直到您回到正确的水平( 或到达底部 ) 。
Node WalkLeft(Node node)
{
if (node == null) return null;
Node leftParent;
switch (node.ChildDirection)
{
case ChildDirection.Root:
return null;
case ChildDirection.TopRight:
return node.Parent.TopLeft;
case ChildDirection.BottomRight:
return node.Parent.BottomLeft;
case ChildDirection.TopLeft:
leftParent = WalkLeft(node.Parent);
return leftParent.TopRight ?? leftParent;
case ChildDirection.BottomLeft:
leftParent = WalkLeft(node.Parent);
return leftParent.BottomLeft ?? leftParent;
}
}
其他方向也同样如此。
x?? y
选择第一个非核值。
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