写一个双向A*

双向A*算法是一种启发式搜索算法，它在搜索过程中从起点和终点同时进行搜索，直到两个搜索路径相遇。以下是一个简单的双向A*算法实现：

假设我们有一个有向图G，节点的表示方式为node，边的表示方式为edge。每个节点都有一个估价函数h(n)，代表从该节点到终点的估计距离。我们需要维护两个数据结构：openSet1和openSet2，分别代表从起点和终点开始的搜索队列。

1. 初始化openSet1和openSet2，将起点和终点分别加入两个队列中。

2. 从openSet1和openSet2中各取出一个节点n1和n2，分别进行扩展。

对于节点n1，遍历它的子节点n1_child，计算从起点到n1_child的实际距离g(n1_child)和估计距离f(n1_child) = g(n1_child) + h(n1_child)。如果n1_child已经在openSet2中，则说明两个搜索路径相遇，搜索结束。如果n1_child不在openSet1中，将n1_child加入openSet1，并记录它的父节点为n1。

对于节点n2，遍历它的父节点n2_parent，计算从终点到n2_parent的实际距离g(n2_parent)和估计距离f(n2_parent) = g(n2_parent) + h(n2_parent)。如果n2_parent已经在openSet1中，则说明两个搜索路径相遇，搜索结束。如果n2_parent不在openSet2中，将n2_parent加入openSet2，并记录它的子节点为n2。

3. 重复步骤2，直到两个搜索路径相遇或者任意一个openSet为空。

下面是一个Python实现代码：

import heapq

def bidirectional_astar(start, goal, neighbors, heuristic):
    # Initialize open sets
    open_set1 = [(heuristic(start, goal), start)]
    open_set2 = [(heuristic(goal, start), goal)]
    closed_set1 = {}
    closed_set2 = {}
    came_from1 = {}
    came_from2 = {}

    while open_set1 and open_set2:
        # Take the node with the lowest f-score from both open sets
        f1, current1 = heapq.heappop(open_set1)
        f2, current2 = heapq.heappop(open_set2)

        if current1 in closed_set2 or current2 in closed_set1:
            # Paths have met, return the final path
            path = reconstruct_path(current1, came_from1, came_from2)
            return path

        # Add the current nodes to their respective closed sets
        closed_set1[current1] = f1
        closed_set2[current2] = f2

        for neighbor in neighbors(current1):
            # Calculate the g-score and f-score for the neighbor
            g = came_from1[current1][0] + distance(current1, neighbor)
            f = g + heuristic(neighbor, goal)

            if neighbor in closed_set1:
                # Ignore the neighbor if it has already been evaluated
                continue

            if (f, neighbor) in open_set1:
                # Update the neighbor's g-score and parent if a better path is found
                index = open_set1.index((f, neighbor))
                if g < came_from1[neighbor][0]:
                    came_from1[neighbor] = (g, current1)
                    open_set1[index] = (f, neighbor)
                    heapq.heapify(open_set1)
            else:
                # Add the neighbor to the open set and record its parent
                came_from1[neighbor] = (g, current1)
                heapq.heappush(open_set1, (f, neighbor))

        for neighbor in neighbors(current2):
            # Calculate the g-score and f-score for the neighbor
            g = came_from2[current2][0] + distance(current2, neighbor)
            f = g + heuristic(goal, neighbor)

            if neighbor in closed_set2:
                # Ignore the neighbor if it has already been evaluated
                continue

            if (f, neighbor) in open_set2:
                # Update the neighbor's g-score and parent if a better path is found
                index = open_set2.index((f, neighbor))
                if g < came_from2[neighbor][0]:
                    came_from2[neighbor] = (g, current2)
                    open_set2[index] = (f, neighbor)
                    heapq.heapify(open_set2)
            else:
                # Add the neighbor to the open set and record its parent
                came_from2[neighbor] = (g, current2)
                heapq.heappush(open_set2, (f, neighbor))

    # No path was found
    return None

def reconstruct_path(current, came_from1, came_from2):
    # Reconstruct the final path by following the parent pointers
    path = [current]
    while current in came_from1:
        current = came_from1[current][1]
        path.append(current)
    path.reverse()
    current = path[-1]
    while current in came_from2:
        current = came_from2[current][1]
        path.append(current)
    return path

该算法的时间复杂度为O(b^(d/2))，其中b是分支因子，d是起点和终点之间的距离。双向A*算法比传统A*算法更快，因为它可以同时从起点和终点开始搜索，在搜索空间中找到相遇点，从而减少了搜索的时间和空间。