我有这行代码测试是否有一个在矩阵表示的迷宫中找到的路径。在我确定是否存在路径之后，如何在最后打印路径。我尝试过堆叠，但我不确定如何继续。
from queue import Queue
maze=open(input())
matrix=maze.readlines()
matrix=[i.strip() for i in matrix]
matrix=[i.split() for i in matrix]
numrows, numcols = len(matrix), len(matrix[0])
q=Queue()
row=0
col=0
q.put((row,col))
while not q.empty():
row, col = q.get()
if col+1 < numcols and matrix[row][col+1] == "0":
q.put((row, col+1))
matrix[row][col+1] = "2"
if row+1 < numrows and matrix[row+1][col] == "0":
q.put((row+1, col))
matrix[row+1][col] = "3"
if 0 <= col-1 and matrix[row][col-1] == "0":
q.put((row, col-1))
matrix[row][col-1] = "4"
if 0 <= row-1 and matrix[row-1][col] == "0":
q.put((row-1, col))
matrix[row-1][col] = "5"
row,col=numrows-1,numcols-1
var=matrix[row][col]
if var=="0":
print ("There is no path.")
else:
print ("There is a path.")
以下是一个示例矩阵：
0 0 0 0 1 0 0 0
0 1 1 0 1 0 1 0
0 1 0 0 1 0 1 0
0 0 0 1 0 0 1 0
0 1 0 1 0 1 1 0
0 0 1 1 0 1 0 0
1 0 0 0 0 1 1 0
0 0 1 1 1 1 0 0
答案 0 :(得分：1)
因为您正在使用BFS来查找路径，所以最终方法不能是从起点到终点的最短路径，实际上，您只能得到哪个点连接到起点的结果。 / p>
因此，要获得精确路径，您可以使用DFS查找路径并使用堆栈来保存所访问的点，或使用Dijkstra找到从起点到终点的最短路径。
这是一个简单的DFS函数，它使用递归，路径用字符2表示
<?php
$host_name = "hostname";
$database = "database";
$user_name = "username";
$password = "pass";
$connect = mysqli_connect($host_name, $user_name, $password, $database);
$select = "SELECT username from userstasker";
$result = $connect->query($select);
$rows = array();
while($r = mysqli_fetch_assoc($result)) {
$rows[] = $r;
}
echo json_encode($rows);
$connect->close();
?>
答案 1 :(得分：1)
这是一种通过从出口向后追踪路径来查找路径的相当简单的方法。您可以通过将移动保存到堆栈来撤消此操作。
为了便于查看，我稍微更改了代码，将N，S，E，W放在矩阵中（北，南，东，西）。要将这些方向字母解码为（行，列）步骤，我使用字典。
from queue import Queue
maze='''\
0 0 0 0 1 0 0 0
0 1 1 0 1 0 1 0
0 1 0 0 1 0 1 0
0 0 0 1 0 0 1 0
0 1 0 1 0 1 1 0
0 0 1 1 0 1 0 0
1 0 0 0 0 1 1 0
0 0 1 1 1 1 0 0
'''
def show(matrix):
for line in matrix:
print(*line)
print()
matrix=maze.splitlines()
matrix=[i.strip() for i in matrix]
matrix=[i.split() for i in matrix]
numrows, numcols = len(matrix), len(matrix[0])
show(matrix)
q=Queue()
row=0
col=0
q.put((row,col))
while not q.empty():
row, col = q.get()
if col+1 < numcols and matrix[row][col+1] == "0":
q.put((row, col+1))
matrix[row][col+1] = "W"
if row+1 < numrows and matrix[row+1][col] == "0":
q.put((row+1, col))
matrix[row+1][col] = "N"
if 0 <= col-1 and matrix[row][col-1] == "0":
q.put((row, col-1))
matrix[row][col-1] = "E"
if 0 <= row-1 and matrix[row-1][col] == "0":
q.put((row-1, col))
matrix[row-1][col] = "S"
row,col=numrows-1,numcols-1
var=matrix[row][col]
show(matrix)
if var=="0":
print ("There is no path.")
exit()
else:
print ("There is a path.")
# Trace the path
step = {'N': (-1, 0), 'S': (1, 0), 'E': (0, 1), 'W': (0, -1)}
while True:
print((row, col), 'go', var)
if row == 0 and col == 0:
break
r, c = step[var]
row += r
col += c
var = matrix[row][col]
print('Home!')
<强>输出
0 0 0 0 1 0 0 0
0 1 1 0 1 0 1 0
0 1 0 0 1 0 1 0
0 0 0 1 0 0 1 0
0 1 0 1 0 1 1 0
0 0 1 1 0 1 0 0
1 0 0 0 0 1 1 0
0 0 1 1 1 1 0 0
E W W W 1 S W W
N 1 1 N 1 S 1 N
N 1 E N 1 S 1 N
N W W 1 S W 1 N
N 1 N 1 S 1 1 N
N W 1 1 S 1 E N
1 N W W W 1 1 N
E N 1 1 1 1 E N
There is a path.
(7, 7) go N
(6, 7) go N
(5, 7) go N
(4, 7) go N
(3, 7) go N
(2, 7) go N
(1, 7) go N
(0, 7) go W
(0, 6) go W
(0, 5) go S
(1, 5) go S
(2, 5) go S
(3, 5) go W
(3, 4) go S
(4, 4) go S
(5, 4) go S
(6, 4) go W
(6, 3) go W
(6, 2) go W
(6, 1) go N
(5, 1) go W
(5, 0) go N
(4, 0) go N
(3, 0) go N
(2, 0) go N
(1, 0) go N
(0, 0) go E
Home!
答案 2 :(得分：1)
作为@程名锐的帖子的变种，DFS的这种实现也将返回它找到的第一个路径。作为一般性建议，如果您不关心最佳路径，DFS通常会比BFS更好地为您服务。
请注意，我们还必须确保不要在圈子中运行，因为此图表不是树。
# for convenience
matrix = [
["0", "0", "0", "0", "1", "0", "0", "0"],
["0", "1", "1", "0", "1", "0", "1", "0"],
["0", "1", "0", "0", "1", "0", "1", "0"],
["0", "0", "0", "1", "0", "0", "1", "0"],
["0", "1", "0", "1", "0", "1", "1", "0"],
["0", "0", "1", "1", "0", "1", "0", "0"],
["1", "0", "0", "0", "0", "1", "1", "0"],
["0", "0", "1", "1", "1", "1", "0", "0"]
]
num_rows = len(matrix)
num_cols = len(matrix[0])
# just to be a bit more flexible, could also be passed as a function argument
goal_state = (num_rows - 1, num_cols - 1)
def dfs(current_path):
# anchor
row, col = current_path[-1]
if (row, col) == goal_state:
return True
# try all directions one after the other
for direction in [(row, col + 1), (row, col - 1), (row + 1, col), (row - 1, col)]:
new_row, new_col = direction
if (0 <= new_row < num_rows and 0 <= new_col < num_cols and # stay in matrix borders
matrix[new_row][new_col] == "0" and # don't run in walls
(new_row, new_col) not in current_path): # don't run in circles
# try new direction
current_path.append((new_row, new_col))
if dfs(current_path): # recursive call
return True
else:
current_path.pop() # backtrack
# result a list of coordinates which should be taken in order to reach the goal
result = [(0, 0)]
if dfs(result):
print("Success!")
print(result)
else:
print("Failure!")
打印：
[(0, 0), (0, 1), (0, 2), (0, 3), (1, 3), (2, 3), (2, 2), (3, 2), (3, 1), (3, 0), (4, 0), (5, 0), (5, 1), (6, 1), (6, 2), (6, 3), (6, 4), (5, 4), (4, 4), (3, 4), (3, 5), (2, 5), (1, 5), (0, 5), (0, 6), (0, 7), (1, 7), (2, 7), (3, 7), (4, 7), (5, 7), (6, 7), (7, 7)]
路径首选项right -> left -> down -> up遇到的DFS的第一个（但肯定不是最短的）正确路径。