问题：
您将获得3条替换规则，每条替换规则之间都可将一个字符串转换为另一个字符串，以及转换所必须使用的步骤数。现在，使用这些规则，您必须打印使用的规则以及到达最终字符串的规则位置。
约束：
每个替换规则介于1到5个字符之间
最大步数s为15
保证可以通过规则的置换将初始字符串转换为最终字符串
样本输入：
AA AB
AB BB
AA B
4 AB AAAB
示例输出：
2 1 BB
3 1 AAB
3 3 AAAA
1 3 AAAB
我尝试了使用BFS在中间相遇中进行双向搜索，该方法对 s =12。DFS在s = 6之前一直工作良好。在1s的时间限制内，是否有任何方法可以解决s = 15的问题？
""" The idea in this code is to implement bidirectional searching with meeting in the middle. Right now, the meeting part is taking around 30s, so I am not concerned about printing the steps right now. """
import math
rule = []
for k in range(3):
rule.append(list(input().split()))
step, start, stop = input().split()
step = int(step)
level = [set() for k in range(step + 1)]
level[0].add(start)
level[step].add(stop)
histup, histdown = {}, {}
histup[start], histdown[stop] = [], []
def bfs(for_start: int, for_end: int, for_step: int, replace_this: int, replace_with: int):
for i in range(for_start, for_end, for_step):
for word in level[i]: # this is the current word we are applying the rules to
for substitution in rule: # this contains the entire substitution
for j in range(len(word) - len(substitution[replace_this]) + 1):
pos = word.find(substitution[replace_this], j)
if pos != -1:
temp = word[:pos]
temp += word[pos:].replace(substitution[replace_this], substitution[replace_with], 1)
"""
if i == math.ceil(step / 2) + 1 and temp in level[i-1]:
#histdown[word].append([rule.index(substitution) + 1, pos + 1, word])
print(temp, '', word)
#print(histup[temp])
#print(histdown[word])
return
"""
level[i + for_step].add(temp)
if for_step == 1:
histup[temp] = histup[word]
histup[temp].append([rule.index(substitution) + 1, pos + 1, temp])
else:
histdown[temp] = histdown[word]
histdown[temp].append([rule.index(substitution) + 1, pos + 1, word])
bfs(0, step-1, 1, 0, 1)
#bfs(0, math.floor(step / 2), 1, 0, 1)
#bfs(step, math.ceil(step / 2), -1, 1, 0)
print(level)
print(len(level[7]))