A*算法求解八数码问题代码python

A*算法是一种启发式搜索算法，常用于路径规划，包括解决八皇后问题这样的回溯问题。在Python中实现A*算法来解决八数码问题，首先需要明确目标状态、起点和启发函数。以下是简化的A*算法伪代码：

import heapq

def a_star_search(board, start, goal):
    frontier = [(0, start)]  # 开放列表，(f_score, state)
    came_from = {}
    cost_so_far = {}

    def heuristic(state):  # 启发函数，计算从当前状态到目标状态的代价估计
        row, col = state
        return max(abs(row - goal[0]), abs(col - goal[1]))

    while frontier:
        current = heapq.heappop(frontier)[1]  # 取出f_score最小的节点
        if current == goal:  # 找到了目标
            break
        for neighbor in get_neighbors(board, current):  # 获取相邻未访问节点
            tentative_g_score = cost_so_far[current] + 1  # 距离加一
            if neighbor not in cost_so_far or tentative_g_score < cost_so_far[neighbor]:
                cost_so_far[neighbor] = tentative_g_score
                priority = tentative_g_score + heuristic(neighbor)  # f_score = g_score + h_score
                heapq.heappush(frontier, (priority, neighbor))
                came_from[neighbor] = current

    solution_path = []
    current = goal
    while current != start:
        solution_path.append(current)
        current = came_from[current]
    solution_path.append(start)  # 添加起始位置
    solution_path.reverse()  # 逆序得到实际路径
    return solution_path

# 辅助函数：获取某个位置的邻居
def get_neighbors(board, pos):
    row, col = pos
    neighbors = [(row+1, col), (row-1, col), (row, col+1), (row, col-1)]
    valid_neighbors = [n for n in neighbors if 0 <= n[0] < len(board) and 0 <= n[1] < len(board[0]) and board[n] == 0]
    return valid_neighbors

# 示例用法
board = [[5,3,0,0,7,0,0,0,0],[6,0,0,1,9,5,0,0,0],[0,9,8,0,0,0,0,6,0],[8,0,0,0,6,0,0,0,3],[4,0,0,8,0,3,0,0,1],[7,0,0,0,2,0,0,0,6],[0,6,0,0,0,0,2,8,0],[0,0,0,4,1,9,0,0,5],[0,0,0,0,8,0,0,7,9]]
start = (0, 0)
goal = (len(board)-1, len(board[0])-1)
path = a_star_search(board, start, goal)