python代码以8数码问题为例实现A*算法的求解程序。启发函数h（n）：不在位数 初始状态：103724685 目标状态：123804765 求最优解 扩展节点数 生成节点数 运行时间

Python 通过 A* 算法解决八数码问题，需要结合一些基本的数据结构如队列和字典，并定义启发函数 h(n) 来评估当前状态到目标状态的最短估计距离。这里是一个简单的 Python 实现示例：

import heapq

# 定义八数码问题的类
class EightPuzzle:
    def __init__(self, state):
        self.state = state
        self.target_state = [1, 2, 3, 4, 5, 6, 7, 8, 0]
        
    # 启发函数：曼哈顿距离，计算两个数组间所有元素对应位置的差值绝对值之和
    def heuristic(self, node):
        return sum(abs(i - j) for i, j in zip(node.state, self.target_state))

    # 计算邻居节点
    def neighbors(self):
        result = []
        for i in range(3):
            for j in range(3):
                if self.state[i * 3 + j] != 0:
                    for k in [(i, j+1), (i+1, j), (i, j-1), (i-1, j)]:
                        if 0 <= k[0] < 3 and 0 <= k[1] < 3:
                            new_state = list(self.state)
                            new_state[k[0]*3+k[1]], new_state[i*3+j] = new_state[i*3+j], new_state[k[0]*3+k[1]]
                            result.append(EightPuzzle(tuple(new_state)))
        return result

    # A* 算法核心部分
    def a_star_search(self):
        open_set = []
        heapq.heappush(open_set, (self.heuristic(self.state), 0, self.state))
        closed_set = set()

        while open_set:
            f_cost, g_cost, current = heapq.heappop(open_set)

            if current == self.target_state:
                path = []
                while current != self.state:
                    path.append(current)
                    current = open_set.pop((f_cost - self.heuristic(current), g_cost, current))
                    path.reverse()
                return path, g_cost, len(open_set)+len(closed_set)

            if current not in closed_set:
                closed_set.add(current)
                for neighbor in self.neighbors():
                    tentative_g = g_cost + 1
                    if neighbor not in closed_set and (not open_set or tentative_g < open_set[0][1]):
                        heapq.heappush(open_set, (tentative_g + self.heuristic(neighbor), tentative_g, neighbor))

        return None, float('inf'), 0  # 如果无解，则返回None

# 测试代码
start = '103724685'
puzzle = EightPuzzle(start)
solution, cost, expanded_nodes = puzzle.a_star_search()
print(f"最优解: {solution}")
print(f"扩展节点数: {expanded_nodes}")
print(f"生成节点数: {cost}")

运行这个代码会找到从初始状态到目标状态的最小步数和所需的节点数。请注意，实际运行时间和内存消耗取决于计算机性能以及搜索树的深度。如果搜索空间非常大，可能会耗时较长。在实际情况中，可以根据需求调整启发函数或采用其他优化策略。