A-star搜索。滑动拼图。这是哪种启发式算法？

A-star搜索（A* search）是一种启发式搜索算法，用于解决问题的最短路径搜索和图搜索问题。它在图中使用了一种启发式评估函数，以估计从当前状态到目标状态的代价。

滑动拼图问题是一个经典的问题，目标是将拼图从初始状态移动到目标状态。下面是使用A-star搜索算法解决滑动拼图问题的代码示例：

import heapq

# 定义滑动拼图问题类
class Puzzle:
    def __init__(self, start, goal):
        self.start = start
        self.goal = goal
    
    # 定义启发式评估函数
    def heuristic(self, node):
        # 计算当前状态与目标状态之间的曼哈顿距离
        distance = 0
        for i in range(len(node)):
            x1, y1 = divmod(node.index(i), 3)
            x2, y2 = divmod(self.goal.index(i), 3)
            distance += abs(x1 - x2) + abs(y1 - y2)
        return distance
    
    # 定义A-star搜索算法
    def astar_search(self):
        # 定义优先队列
        open_list = []
        # 将起始状态加入优先队列中，使用启发式估值作为优先级
        heapq.heappush(open_list, (self.heuristic(self.start), self.start))
        # 定义已访问状态集合
        closed_list = set()
        # 定义父状态字典，用于记录每个状态的父状态
        parents = {}
        # 定义代价字典，用于记录每个状态的代价
        costs = {self.start: 0}
        
        while open_list:
            # 从优先队列中取出优先级最高的状态
            current = heapq.heappop(open_list)[1]
            # 如果当前状态为目标状态，停止搜索
            if current == self.goal:
                break
            # 将当前状态加入已访问状态集合中
            closed_list.add(current)
            # 获取当前状态的邻居状态
            neighbors = self.get_neighbors(current)
            
            for neighbor in neighbors:
                # 计算邻居状态的代价
                new_cost = costs[current] + 1
                # 如果邻居状态未被访问过或新的代价更小
                if neighbor not in costs or new_cost < costs[neighbor]:
                    # 更新邻居状态的代价
                    costs[neighbor] = new_cost
                    # 更新邻居状态的父状态
                    parents[neighbor] = current
                    # 将邻居状态加入优先队列中，使用启发式估值和代价之和作为优先级
                    heapq.heappush(open_list, (new_cost + self.heuristic(neighbor), neighbor))
        
        # 从目标状态开始，依次追溯父状态，得到最短路径
        path = []
        current = self.goal
        while current != self.start:
            path.append(current)
            current = parents[current]
        path.append(self.start)
        path.reverse()
        
        return path
    
    # 获取当前状态的邻居状态
    def get_neighbors(self, state):
        neighbors = []
        # 获取空格的位置
        blank_index = state.index(0)
        # 上移
        if blank_index >= 3:
            new_state = state[:]
            new_state[blank_index], new_state[blank_index - 3] = new_state[blank_index - 3], new_state[blank_index]
            neighbors.append(new_state)
        # 下移
        if blank_index < 6:
            new_state = state[:]
            new_state[blank_index], new_state[blank_index + 3] = new_state[blank_index + 3], new_state[blank_index]
            neighbors.append(new_state)
        # 左移
        if blank_index % 3 != 0:
            new_state = state[:]
            new_state[blank_index], new_state[blank_index - 1] = new_state[blank_index - 1], new_state[blank_index]
            neighbors.append(new_state)
        #