一个魔方积木的还原算法
魔方积木如图：
说明：需要通过转动拼成3*3的立方体，接口处可以转动。
# 运行
python3 cube.py
# 共有两种不同的解法
# 输出结果如下：
**************************************** 1
(0, 0, 0) -> z+2 -> [(0, 0, 1), (0, 0, 2)]
(0, 0, 2) -> x+2 -> [(1, 0, 2), (2, 0, 2)]
(2, 0, 2) -> z-2 -> [(2, 0, 1), (2, 0, 0)]
(2, 0, 0) -> y+2 -> [(2, 1, 0), (2, 2, 0)]
(2, 2, 0) -> z+1 -> [(2, 2, 1)]
(2, 2, 1) -> y-1 -> [(2, 1, 1)]
(2, 1, 1) -> z+1 -> [(2, 1, 2)]
(2, 1, 2) -> x-2 -> [(1, 1, 2), (0, 1, 2)]
(0, 1, 2) -> z-2 -> [(0, 1, 1), (0, 1, 0)]
(0, 1, 0) -> y+1 -> [(0, 2, 0)]
(0, 2, 0) -> x+1 -> [(1, 2, 0)]
(1, 2, 0) -> y-2 -> [(1, 1, 0), (1, 0, 0)]
(1, 0, 0) -> z+1 -> [(1, 0, 1)]
(1, 0, 1) -> y+2 -> [(1, 1, 1), (1, 2, 1)]
(1, 2, 1) -> x-1 -> [(0, 2, 1)]
(0, 2, 1) -> z+1 -> [(0, 2, 2)]
(0, 2, 2) -> x+2 -> [(1, 2, 2), (2, 2, 2)]
**************************************** 2
(0, 0, 0) -> z+2 -> [(0, 0, 1), (0, 0, 2)]
(0, 0, 2) -> y+2 -> [(0, 1, 2), (0, 2, 2)]
(0, 2, 2) -> z-2 -> [(0, 2, 1), (0, 2, 0)]
(0, 2, 0) -> x+2 -> [(1, 2, 0), (2, 2, 0)]
(2, 2, 0) -> z+1 -> [(2, 2, 1)]
(2, 2, 1) -> x-1 -> [(1, 2, 1)]
(1, 2, 1) -> z+1 -> [(1, 2, 2)]
(1, 2, 2) -> y-2 -> [(1, 1, 2), (1, 0, 2)]
(1, 0, 2) -> z-2 -> [(1, 0, 1), (1, 0, 0)]
(1, 0, 0) -> x+1 -> [(2, 0, 0)]
(2, 0, 0) -> y+1 -> [(2, 1, 0)]
(2, 1, 0) -> x-2 -> [(1, 1, 0), (0, 1, 0)]
(0, 1, 0) -> z+1 -> [(0, 1, 1)]
(0, 1, 1) -> x+2 -> [(1, 1, 1), (2, 1, 1)]
(2, 1, 1) -> y-1 -> [(2, 0, 1)]
(2, 0, 1) -> z+1 -> [(2, 0, 2)]
(2, 0, 2) -> y+2 -> [(2, 1, 2), (2, 2, 2)]
说明：
(0, 0, 0) -> z+2 -> [(0, 0, 1), (0, 0, 2)]这个的意思是，当前的起点坐标为(0,0,0)，做操作z+2（z轴增加2，说明边长为3的就是增加2），即边(0,0,0)->(0,0,2)，可以理解为将该边进行染色，被染色的点为[(0, 0, 1), (0, 0, 2)]
原点(0, 0, 0)初始状态就已经染色了。