在路径查找时我是初学者，虽然我理解A *的基本思想，但我仍然不明白为什么，当回溯时，实现不会卡在两个最新节点之间的循环中参观。
为了更清楚，我一直在查看来自here的代码（我将要复制并粘贴，以防链接死亡）：
class Node():
"""A node class for A* Pathfinding"""
def __init__(self, parent=None, position=None):
self.parent = parent
self.position = position
self.g = 0
self.h = 0
self.f = 0
def __eq__(self, other):
return self.position == other.position
def astar(maze, start, end):
"""Returns a list of tuples as a path from the given start to the given end in the given maze"""
# Create start and end node
start_node = Node(None, start)
start_node.g = start_node.h = start_node.f = 0
end_node = Node(None, end)
end_node.g = end_node.h = end_node.f = 0
# Initialize both open and closed list
open_list = []
closed_list = []
# Add the start node
open_list.append(start_node)
# Loop until you find the end
while len(open_list) > 0:
# Get the current node
current_node = open_list[0]
current_index = 0
for index, item in enumerate(open_list):
if item.f < current_node.f:
current_node = item
current_index = index
# Pop current off open list, add to closed list
open_list.pop(current_index)
closed_list.append(current_node)
# Found the goal
if current_node == end_node:
path = []
current = current_node
while current is not None:
path.append(current.position)
current = current.parent
return path[::-1] # Return reversed path
# Generate children
children = []
for new_position in [(0, -1), (0, 1), (-1, 0), (1, 0), (-1, -1), (-1, 1), (1, -1), (1, 1)]: # Adjacent squares
# Get node position
node_position = (current_node.position[0] + new_position[0], current_node.position[1] + new_position[1])
# Make sure within range
if node_position[0] > (len(maze) - 1) or node_position[0] < 0 or node_position[1] > (len(maze[len(maze)-1]) -1) or node_position[1] < 0:
continue
# Make sure walkable terrain
if maze[node_position[0]][node_position[1]] != 0:
continue
# Create new node
new_node = Node(current_node, node_position)
# Append
children.append(new_node)
# Loop through children
for child in children:
# Child is on the closed list
for closed_child in closed_list:
if child == closed_child:
continue
# Create the f, g, and h values
child.g = current_node.g + 1
child.h = ((child.position[0] - end_node.position[0]) ** 2) + ((child.position[1] - end_node.position[1]) ** 2)
child.f = child.g + child.h
# Child is already in the open list
for open_node in open_list:
if child == open_node and child.g > open_node.g:
continue
# Add the child to the open list
open_list.append(child)
def main():
maze = [[0, 0, 0, 0, 1, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 1, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 1, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 1, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 1, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 1, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 1, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 1, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0]]
start = (0, 0)
end = (7, 6)
path = astar(maze, start, end)
print(path)
if __name__ == '__main__':
main()
在这种特殊情况下，这看起来很容易，但如果“迷宫”是这样的：
maze =
[[0, 0, 0, 0, 1, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 1, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 1, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 1, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 1, 1, 1, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 1, 0, 0, 0],
[0, 0, 0, 0, 1, 1, 1, 0, 0, 0],
[0, 0, 0, 0, 1, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 1, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0]]
当被检查的节点是(5, 5)时，有一个死胡同，所以算法应该“回溯”到节点(3, 5)并从那里下来。
我的问题是我真的不明白这是怎么回事。返回一个节点并检查邻居意味着再次检查所有内容，它将返回(5, 5)。
这不会发生，因为实现工作得很好，但我似乎真的无法掌握如何。任何提示？
关键是当运行Dijkstra或A *时，每个节点最多访问一次。
这是因为每次访问节点时（在我们从队列中弹出它之后），我们都会“标记”此节点已被访问。在您给出的实现中，通过将节点添加到closed_list来进行标记：
closed_list.append(current_node)
现在节点是closed_list的事实将允许在将节点推送到队列之前检查节点是否已经被访问过。这在代码中进一步完成（笨拙地）：
# Child is on the closed list
for closed_child in closed_list:
if child == closed_child:
continue
这种机制对于确保算法终止（Dijkstra或A *）以及其复杂性在O(n.log n)中是必不可少的。
但是在大多数实现中，你不会看到closed_list，而是一个与每个节点相关联的visited布尔值，或者一种颜色（如果没有访问则为white，如果已访问则为green）。它们在终止保证方面都是等价的（不一定是在性能方面，因为列表查找可能会花费O(n)）。