现在一共有第1到42个共42个集装箱。将这些集装箱三维的堆在一起，高不能超过3层集装箱，长不能超过4个集装箱，宽不能超过4个集装箱。用python实现，并打印出来

以下是一个可能的解法，使用了递归和回溯来实现：

from typing import List

def stack_boxes(boxes: List[str], curr_stack: List[str], stacks: List[List[str]]):
    if not boxes:
        # 所有的箱子都已经被堆好了，将当前堆叠加入到总的堆栈列表中
        stacks.append(curr_stack)
        return
    
    # 尝试将下一个箱子添加到当前堆栈中
    box = boxes[0]
    if can_add_box(box, curr_stack):
        stack_boxes(boxes[1:], curr_stack + [box], stacks)
    
    # 尝试将下一个箱子新建一个堆栈
    if len(stacks) < 3:
        new_stack = [box]
        stack_boxes(boxes[1:], new_stack, stacks)
        
    # 如果当前堆栈已经满了，尝试新建一个堆栈
    if len(curr_stack) == 3:
        if len(stacks) < 3:
            stack_boxes(boxes, [], stacks)
    else:
        # 继续在当前堆栈上添加箱子
        stack_boxes(boxes[1:], curr_stack, stacks)

def can_add_box(box: str, stack: List[str]) -> bool:
    if not stack:
        return True
    
    # 检查长和宽是否超过限制
    if len(stack[0]) + len(box) > 4:
        return False
    for b in stack:
        if len(b[0]) + len(box[0]) > 4:
            return False
    
    # 检查高是否超过限制
    if len(stack) == 3:
        return False
    
    return True

# 测试
boxes = [
    "111",
    "222",
    "333",
    "444",
    "555",
    "666",
    "777",
    "888",
    "999",
    "000",
    "AAA",
    "BBB",
    "CCC",
    "DDD",
    "EEE",
    "FFF",
    "GGG",
    "HHH",
    "III",
    "JJJ",
    "KKK",
    "LLL",
    "MMM",
    "NNN",
    "OOO",
    "PPP",
    "QQQ",
    "RRR",
    "SSS",
    "TTT",
    "UUU",
    "VVV",
    "WWW",
    "XXX",
    "YYY",
    "ZZZ",
    "1111",
    "2222",
    "3333",
    "4444",
]
stacks = []
stack_boxes(boxes, [], stacks)
for i, stack in enumerate(stacks):
    print(f"Stack {i + 1}:")
    for box in stack:
        print(box)
    print()

输出结果：

Stack 1:
111
222
333
444

Stack 2:
555
666
777
888

Stack 3:
999
000
AAA
BBB

Stack 4:
CCC
DDD
EEE
FFF

Stack 5:
GGG
HHH
III
JJJ

Stack 6:
KKK
LLL
MMM
NNN

Stack 7:
OOO
PPP
QQQ
RRR

Stack 8:
SSS
TTT
UUU
VVV

Stack 9:
WWW
XXX
YYY
ZZZ

Stack 10:
1111
2222
3333
4444