A 寻路算法常见陷阱：节点探索中断问题诊断与修正

def AStar(start_node, end_node, graph, heuristic):
    openSet = PriorityQueue() # 假设PriorityQueue已正确实现
    openSet.enqueue(0, start_node)

    infinity = float("inf")

    gCost = {} # 存储从起点到各节点的实际代价
    fCost = {} # 存储从起点到各节点的总估计代价
    cameFrom = {} # 存储路径回溯信息

    for node in graph:
        gCost[node] = infinity
        fCost[node] = infinity
    gCost[start_node] = 0
    fCost[start_node] = heuristic(start_node, end_node)

    while not openSet.isEmpty():
        current = openSet.dequeue()

        if current == end_node:
            # RetracePath(cameFrom, end_node) # 路径回溯逻辑
            return True # 找到路径

        # 错误所在：这里始终探索的是start_node的邻居
        for neighbour in find_neighbors(start_node, graph): # <-- 错误点
            tempGCost = gCost[current] + 1 # 假设每一步代价为1

            if tempGCost < gCost[neighbour]:
                cameFrom[neighbour] = current
                gCost[neighbour] = tempGCost
                fCost[neighbour] = tempGCost + heuristic(neighbour, end_node)

                if not openSet.contains(neighbour): # 假设PriorityQueue支持contains方法
                    openSet.enqueue(fCost[neighbour], neighbour)
    return False # 未找到路径

def find_neighbors(node, graph):
    x, y = node
    neighbors = []
    # 假设graph是一个包含所有可行节点的集合或字典
    # 示例中只考虑了上下左右四个方向
    possible_neighbors = [(x + 1, y), (x - 1, y), (x, y + 1), (x, y - 1)]

    for neighbor_coord in possible_neighbors:
        if neighbor_coord in graph: # 检查邻居是否在图中（即是否可通行）
            neighbors.append(neighbor_coord)
    return neighbors

import heapq # 使用Python内置的heapq模块实现优先队列

class PriorityQueue:
    def __init__(self):
        self._queue = []
        self._index = 0

    def enqueue(self, priority, item):
        # heapq是小顶堆，存储 (priority, index, item) 以处理相同priority的情况
        heapq.heappush(self._queue, (priority, self._index, item))
        self._index += 1

    def dequeue(self):
        return heapq.heappop(self._queue)[2] # 返回item

    def isEmpty(self):
        return len(self._queue) == 0

    # 注意：PriorityQueue通常不直接提供O(1)或O(logN)的contains方法
    # 对于A*，我们通过gCost和fCost字典来判断节点是否已被访问或更新
    # 这里的contains实现是为了演示，实际中通常不这样用或需要更复杂的实现
    def contains(self, item):
        for _, _, existing_item in self._queue:
            if existing_item == item:
                return True
        return False # 实际应用中，更高效的方法是检查gCost是否为无穷大

def AStar(start_node, end_node, graph, heuristic):
    openSet = PriorityQueue()
    openSet.enqueue(0, start_node)

    infinity = float("inf")

    gCost = {node: infinity for node in graph} # 从起点到当前节点的实际代价
    fCost = {node: infinity for node in graph} # 从起点到当前节点的总估计代价
    cameFrom = {} # 存储路径回溯信息

    gCost[start_node] = 0
    fCost[start_node] = heuristic(start_node, end_node)

    while not openSet.isEmpty():
        current = openSet.dequeue()

        if current == end_node:
            return RetracePath(cameFrom, end_node) # 找到路径，回溯并返回

        # 修正点：现在探索的是current节点的邻居
        for neighbour in find_neighbors(current, graph): # <-- 修正点
            # 假设每一步代价为1，如果不同，需要从graph中获取边的权重
            tempGCost = gCost[current] + 1

            if tempGCost < gCost[neighbour]: # 发现更短的路径
                cameFrom[neighbour] = current
                gCost[neighbour] = tempGCost
                fCost[neighbour] = tempGCost + heuristic(neighbour, end_node)

                # 只有当邻居不在openSet中，或者找到了更优的路径时才更新/添加
                # 注意：如果PriorityQueue不支持高效的update操作，
                # 常见做法是直接添加新项，让旧的、较差的项留在队列中，
                # 当它被取出时，由于gCost[neighbour]已更新，会被忽略。
                if not openSet.contains(neighbour): # 简化判断，实际可优化
                    openSet.enqueue(fCost[neighbour], neighbour)
    return None # 未找到路径

def RetracePath(cameFrom, current):
    path = [current]
    while current in cameFrom:
        current = cameFrom[current]
        path.append(current)
    return path[::-1] # 反转路径，使其从起点到终点

# 示例启发式函数（曼哈顿距离）
def heuristic(node_a, node_b):
    return abs(node_a[0] - node_b[0]) + abs(node_a[1] - node_b[1])

# 示例图数据 (假设是一个2D网格，graph包含所有可通行坐标)
# 例如：graph = {(0,0), (0,1), (1,0), (1,1), ...}
# 为了简单，这里假设一个4x4的网格，所有节点都可通行
example_graph = set()
for x in range(4):
    for y in range(4):
        example_graph.add((x, y))

# 示例用法
start = (0, 0)
goal = (3, 3)
path = AStar(start, goal, example_graph, heuristic)
if path:
    print(f"找到路径: {path}")
else:
    print("未找到路径")

# 另一个示例，模拟问题中的输出
enemy_coords = (6, 2)
player_coords = (10, 2)
# 假设一个更大的图，包含这些坐标
large_graph = set()
for x in range(0, 15):
    for y in range(0, 5):
        large_graph.add((x, y))

path_to_player = AStar(enemy_coords, player_coords, large_graph, heuristic)
if path_to_player:
    print(f"敌人到玩家的路径: {path_to_player}")
else:
    print("敌人未能找到到达玩家的路径")