📅  最后修改于: 2023-12-03 15:28:48.765000             🧑  作者: Mango
这是一道典型的计算机科学问题,涉及图论和搜索算法。问题描述如下:
给定一个包含n个门和n个钥匙的房间,每个门和钥匙都有唯一的编号。你需要找到最短的路径,使得每个门都能被打开。每个门只能用与之对应的钥匙打开。你可以同时拿走若干把钥匙,并回溯到已经访问过的门。如果没有解决方案,则输出"No solution"。
以下是解决该问题的一种算法:
以下是该算法的代码实现,使用Python语言:
from collections import deque
# 构建图
def build_graph(doors, keys):
graph = {}
n = len(doors)
for i in range(n):
door = doors[i]
key = keys[i]
graph[door] = (door+'i', door+'o')
graph[key] = (key+'i', key+'o')
# 添加门和钥匙之间的连接关系
graph[door+'i'] = (door, key+'i')
graph[door+'o'] = (door, key+'o')
# 添加钥匙和门之间的连接关系
graph[key+'i'] = (key, door+'i')
graph[key+'o'] = (key, door+'o')
return graph
# 搜索路径
def search(doors, keys):
graph = build_graph(doors, keys)
start = doors[0]
visited = {start}
q = deque([(start, 0, [])])
while q:
node, dist, path = q.popleft()
if len(path) == len(doors):
return path, dist
for neighbor in graph[node]:
if neighbor not in visited:
visited.add(neighbor)
q.append((neighbor, dist+1, path+[node]))
return None
# 主函数
def main():
n = int(input())
doors = input().split()
keys = input().split()
path, dist = search(doors, keys)
if path:
print(' '.join(path))
print(dist)
else:
print('No solution')
以上代码通过构建无向图和BFS搜索算法,实现了寻找最短路径的功能。该算法的时间复杂度为O(n^2),空间复杂度为O(n)。该算法不仅可用于解决门和钥匙问题,也可以用于解决其他图论问题。