白给题，查看html源代码发现每个点位的方块有'top', 'bottom', 'left', 'right'，表示该点上下左右不能走。直接使用正则表达式提取每个点的坐标及其对应的状态，用1+2+4+8存储，然后以图上每一个点作为初始点深度优先搜索即可，每次更新最大值maxn
特别注意此处dfs会爆栈，需要手动扩栈
交了一次发现错误，意识到可能形成了环，所以还需要+1，结果应该为4056
Maze文件地址Maze.html
import re
import sys
sys.setrecursionlimit(100000)
wyx = [[0 for i in range(100)] for i in range(100)]
vis = [[0 for i in range(100)] for i in range(100)]
with open("Maze.html") as file_object:
str1 = file_object.read()
print(str1)
str2 = re.findall(r"<td (.+?);\"> </td>", str1)
for i in str2:
a = re.findall(r"id=\"(.+?)-", i)
b = re.findall(r"-(.+?)\"", i)
if 'top' in i:
wyx[int(a[0])][int(b[0])] += 1
if 'bottom' in i:
wyx[int(a[0])][int(b[0])] += 2
if 'left' in i:
wyx[int(a[0])][int(b[0])] += 4
if 'right' in i:
wyx[int(a[0])][int(b[0])] += 8
maxn = 0
def judge(x, y):
if x < 0 or x > 99:
return False
if y < 0 or y > 99:
return False
if vis[x][y] == 1:
return False
return True
def dfs(x, y, step):
global maxn
maxn = max(maxn, step)
v = wyx[x][y]
vis[x][y] = 1
if v >= 8:
v -= 8
else:
if judge(x, y + 1):
dfs(x, y + 1, step + 1)
if v >= 4:
v -= 4
else:
if judge(x, y - 1):
dfs(x, y - 1, step + 1)
if v >= 2:
v -= 2
else:
if judge(x + 1, y):
dfs(x + 1, y, step + 1)
if v >= 1:
v -= 1
else:
if judge(x - 1, y):
dfs(x - 1, y, step + 1)
vis[x][y] = 0
for i in range(100):
for j in range(100):
dfs(i, j, 0)
print maxn
import hashlib
print hashlib.md5('4056').digest().encode('hex')