python解华容道问题的程序

下面是一个简单的Python程序，可以解决华容道问题。程序使用广度优先搜索算法，遍历所有可能的移动序列，直到找到解决方案。

from collections import deque

# 定义华容道类
class HuaRongDao:
    def __init__(self, board):
        self.board = board
        self.empty_pos = self.get_empty_pos()

    # 获取空白位置
    def get_empty_pos(self):
        for i in range(5):
            for j in range(4):
                if self.board[i][j] == 0:
                    return (i, j)

    # 移动棋子
    def move(self, pos1, pos2):
        i1, j1 = pos1
        i2, j2 = pos2
        self.board[i1][j1], self.board[i2][j2] = self.board[i2][j2], self.board[i1][j1]

    # 获取所有可能移动的位置
    def get_moves(self):
        moves = []
        i, j = self.empty_pos
        if i > 0:
            moves.append((i-1, j))
        if i < 4:
            moves.append((i+1, j))
        if j > 0:
            moves.append((i, j-1))
        if j < 3:
            moves.append((i, j+1))
        return moves

    # 获取下一步棋盘状态
    def get_next_board(self, move):
        next_board = [[self.board[i][j] for j in range(4)] for i in range(5)]
        self.move(self.empty_pos, move)
        next_hrdao = HuaRongDao(next_board)
        next_hrdao.empty_pos = move
        return next_hrdao

    # 判断是否达到目标状态
    def is_solved(self):
        return self.board[3][1] == 1 and self.board[3][2] == 2

# 广度优先搜索
def bfs(initial_hrdao):
    queue = deque([(initial_hrdao, [])])
    visited = set()
    while queue:
        hrdao, moves = queue.popleft()
        if hrdao.is_solved():
            return moves
        visited.add(str(hrdao.board))
        for move in hrdao.get_moves():
            next_hrdao = hrdao.get_next_board(move)
            if str(next_hrdao.board) not in visited:
                queue.append((next_hrdao, moves + [move]))

# 棋盘状态
board = [
    [1, 1, 0, 2],
    [1, 1, 0, 2],
    [3, 4, 4, 5],
    [6, 7, 8, 5],
    [6, 9, 9, 10]
]

# 初始华容道
initial_hrdao = HuaRongDao(board)

# 解题
moves = bfs(initial_hrdao)
print(moves)

你可以将程序复制到Python编辑器中运行，得到华容道问题的解决方案。注意，程序中的棋盘状态可以根据实际情况进行修改。