我正在努力在Bloxorz游戏上实现 a-Star 算法。其目标是使用 1 x 1 x 2 块到达终点。我实现了算法,但它是不一致的。有时它没有给出最短的解决方案。例如:

maze = ['00011111110000',
        '00011111110000',
        '11110000011100',
        '11100000001100',
        '11100000001100',
        '1S100111111111',
        '11100111111111',
        '000001E1001111',
        '00000111001111']

对于这个迷宫，我的实现给出了这个结果：

U，L，U，R，R，U，R，R，R，R，R，R，D，R，D，D，L，L，L，D，R，D，L，U，R，U，L，D

它有 29 个动作。但是有一个更短的解决方案有 28 个动作：

U，L，U，R，R，U，R，R，R，R，R，R，D，R，D，D，D，R，U，L，L，L，L，L，L，L，D

这是我的实现，完整的代码是here，我能做些什么呢？

cl Node:
    def __init__(self,parent:'Node', node_type:str, x1:int, y1:int, x2:int, y2:int, direction:str=''):
        self.parent = parent
        self.node_type = node_type
        self.g = 0
        self.h = 0
        self.f = 0
        self.x1 = x1
        self.y1 = y1
        self.x2 = x2
        self.y2 = y2
        self.visited = False
        self.direction = direction
    def get_positions(self) -> tuple:
        return (self.x1, self.y1, self.x2, self.y2)
    def __eq__(self, other):
        if type(other) is Node:
            return self.x1 == other.x1 and self.y1 == other.y1 and self.x2 == other.x2 and self.y2 == other.y2
        elif type(other) is tuple:
            return self.x1 == other[0] and self.y1 == other[1] and self.x2 == other[2] and self.y2 == other[3]
        else:
            return False
    def __lt__(self, other:'Node'):
        return self.f < other.f
def aStar(start:Node, end:Node, grid:List[List[str]]) -> List[tuple]:
    open_list = []
    closed_list = []
    heapq.heappush(open_list, start)
    while open_list:
        current:Node = heapq.heappop(open_list)
        if current == end:
            return reconstruct_path(current)
        closed_list.append(current)
        for neighbor in get_neighbors(current, grid):
            if neighbor not in closed_list:
                neighbor.g = current.g + 1
                neighbor.h = get_heuristic(neighbor, end)
                neighbor.f = neighbor.g + neighbor.h
                if neighbor not in open_list:
                    heapq.heappush(open_list, neighbor)
    return []
def reconstruct_path(current:Node) -> List[tuple]:
    path = []
    while current.parent is not None:
        path.append(current.direction)
        current = current.parent
    return ''.join(path[::-1])
def get_heuristic(current:Node, end:Node) -> int:
    return max(abs(current.x2 - end.x1), abs(current.y2 - end.y1))