我正在尝试在python中实现广度优先搜索。我正试图找到一条通过网格的路径，从一个方格开始，找到一条通向球门方的路径。整个网格中存在由字符“o”标记的障碍。 “图”指的是节点的二维数组，一个简单的类：
# Node class
class Node:
def __init__(self, val, pos):
self.val = val
self.pos = pos
self.visited = False
def __str__(self):
return "%s" % self.val
我意识到这不是BFS最干净的实现 - 我没有太多使用python的经验，所以有时我不得不使用重复代码，因为我不确定python如何处理某些下面的指针局部变量。无论如何，这个BFS无限循环，我无法弄清楚为什么。任何帮助，将不胜感激！
边缘基本上是一个队列，在移动更深层次之前，按左，上，右和下的顺序检查每个节点的相邻正方形。
# Breadth First Search
def bfs(arena, start):
# fringe implemented as a FIFO list (behaves like a queue)
fringe = []
fringe.append([start])
start.visited = True
while fringe:
# get the first path from the fringe queue
path = fringe.pop(0)
print "pop!"
# get the last node from the path
node = path[-1]
# goal check
if node.val == 'g':
print "PATH FOUND!!"
return path
# get all adjacent nodes, construct a new path and push it to the fringe queue
pos = node.pos
# get left node first
if pos[1]-1>=0:
neighbor = graph[pos[0]][pos[1]-1]
newPath = path[:]
if neighbor.val == 'o':
neighbor.visited = True
graph[pos[0]][pos[1]-1].visited = True
if neighbor is not neighbor.visited:
neighbor.visited = True
graph[pos[0]][pos[1]-1].visited = True
newPath.append(neighbor)
fringe.append(newPath)
print "left node added!"
# get node above current node
if pos[0]-1>=0:
neighbor = graph[pos[0]-1][pos[1]]
newPath = path[:]
if neighbor.val == 'o':
neighbor.visited = True
graph[pos[0-1]][pos[1]].visited = True
if neighbor is not neighbor.visited:
neighbor.visited = True
graph[pos[0-1]][pos[1]].visited = True
newPath.append(neighbor)
fringe.append(newPath)
print "top noded added!"
# get node to the right of current node
if pos[1]+1 < columns:
neighbor = graph[pos[0]][pos[1]+1]
newPath = path[:]
if neighbor.val == 'o':
neighbor.visited = True
graph[pos[0]][pos[1]+1].visited = True
if neighbor is not neighbor.visited:
neighbor.visited = True
graph[pos[0]][pos[1]+1].visited = True
newPath.append(neighbor)
fringe.append(newPath)
print "right node added!"
# get node below current node
if pos[0]+1 < rows:
neighbor = graph[pos[0]+1][pos[1]]
newPath = path[:]
if neighbor.val == 'o':
neighbor.visited = True
graph[pos[0]+1][pos[1]].visited = True
if neighbor is not neighbor.visited:
neighbor.visited = True
graph[pos[0]+1][pos[1]].visited = True
newPath.append(neighbor)
fringe.append(newPath)
print "node below added!"
答案 0 :(得分：2)
这是工作代码，请仔细阅读评论。
from pprint import pprint
# pos is (row, column), not (x, y)
class Node:
def __init__(self, val, pos):
self.val = val
# Position info is stored here and ALSO as index in graph -
# this is a form of data duplication, which is evil!
self.pos = pos
self.visited = False
def __repr__(self):
# nice repr for pprint
return repr(self.pos)
# You had mistake here, "arena" instead of "graph"
# Before posting questions on stackoverflow, make sure your examples
# at least produce some incorrect result, not just crash.
# Don't make people fix syntax errors for you!
def bfs(graph, start):
fringe = [[start]]
# Special case: start == goal
if start.val == 'g':
return [start]
start.visited = True
# Calculate width and height dynamically. We assume that "graph" is dense.
width = len(graph[0])
height = len(graph)
# List of possible moves: up, down, left, right.
# You can even add chess horse move here!
moves = [(-1, 0), (1, 0), (0, -1), (0, 1)]
while fringe:
# Print fringe at each step
pprint(fringe)
print('')
# Get first path from fringe and extend it by possible moves.
path = fringe.pop(0)
node = path[-1]
pos = node.pos
# Using moves list (without all those if's with +1, -1 etc.) has huge benefit:
# moving logic is not duplicated. It will save you from many silly errors.
# The example of such silly error in your code:
# graph[pos[0-1]][pos[1]].visited = True
# ^^^
# Also, having one piece of code instead of four copypasted pieces
# will make algorithm much easier to change (e.g. if you want to move diagonally).
for move in moves:
# Check out of bounds. Note that it's the ONLY place where we check it. Simple and reliable!
if not (0 <= pos[0] + move[0] < height and 0 <= pos[1] + move[1] < width):
continue
neighbor = graph[pos[0] + move[0]][pos[1] + move[1]]
if neighbor.val == 'g':
return path + [neighbor]
elif neighbor.val == 'o' and not neighbor.visited:
# In your original code there was a line:
# if neighbor is not neighbor.visited:
# which was completely wrong. Read about "is" operator.
neighbor.visited = True
fringe.append(path + [neighbor]) # creates copy of list
raise Exception('Path not found!')
if __name__ == '__main__':
# Graph in convenient form: 0 is empty, 1 is wall, 2 is goal.
# Note that you can have multiple goals.
graph = [
[0, 1, 0, 1],
[0, 0, 0, 0],
[0, 1, 0, 1],
[0, 0, 2, 1]
]
# Transform int matrix to Node matrix.
TRANSLATE = {0: 'o', 1: 'x', 2: 'g'}
graph = [[Node(TRANSLATE[x], (i, j)) for j, x in enumerate(row)] for i, row in enumerate(graph)]
# Find path
try:
path = bfs(graph, graph[0][0])
print("Path found: {!r}".format(path))
except Exception as ex:
# Learn to use exceptions. In your original code, "no path" situation
# is not handled at all!
print(ex)