leetcode(28)BFS系列题目
2368. 受限条件下可到达节点的数目
步骤：
1.根据边建好无向图
[[1, 4, 5], [0, 2, 3], [1], [1], [0], [0, 6], [5]]
2.初始化访问过的节点visited里已经有受限节点了就不会再去访问，并且是set类型的，即不能重复访问
3.q里是当前能够到达的节点u，找到与u相连但是还没有访问（也没有被限制）的节点v
class Solution:
def reachableNodes(self, n: int, edges: List[List[int]], restricted: List[int]) -> int:
graph = [[] for _ in range(n)]
for u, v in edges:
graph[u].append(v)
graph[v].append(u)
print(graph)
visited = set(restricted)
visited.add(0)
q = [0]
res = 1
while q:
u = q.pop()
for v in graph[u]:
if v not in visited:
visited.add(v)
q.append(v)
res += 1
return res
864. 获取所有钥匙的最短路径
没有进行状态压缩，state 记录当前收集过的钥匙
class Solution:
def shortestPathAllKeys(self, grid: List[str]) -> int:
keys, locks = 'abcdef', 'ABCDEF'
cnt, res = 0, 0
m, n = len(grid), len(grid[0])
for i in range(m):
for j in range(n):
if grid[i][j] in keys:
cnt += 1
if grid[i][j] == '@':
begin = (i, j)
q = deque([(begin[0], begin[1], '')])
vis = set()
dire = [(-1,0),(1,0),(0,1),(0,-1)]
while q:
for _ in range(len(q)):
i, j, state = q.popleft() # 注意是left
if (i, j, state) in vis:
continue
if len(state) == cnt:
return res
vis.add((i, j, state))
for x, y in dire:
newi = i + x
newj = j + y
if not (0 <= newi < m and 0 <= newj < n and grid[newi][newj] != '#'):
continue # 1.此路不通
if grid[newi][newj] in locks and grid[newi][newj].lower() in state:
q.append((newi, newj, state)) # 2.遇到锁但是有钥匙，可通
elif grid[newi][newj] in keys:
if grid[newi][newj] in state:
q.append((newi, newj, state)) # 3.遇到收集过的钥匙
else:
q.append((newi, newj, state + grid[newi][newj])) # 4.遇到没收集过的钥匙
elif grid[newi][newj] in '.@': # 5.空房间可通
q.append((newi, newj, state))
res += 1
return -1
815. 公交路线
class Solution:
def numBusesToDestination(self, routes: List[List[int]], source: int, target: int) -> int:
# 每个车站可以乘坐第几辆公交车
stations = defaultdict(set)
for i, stops in enumerate(routes):
for stop in stops:
stations[stop].add(i)
# print(stations)
# 每辆公交车可以到达哪些车站，转成set
routes = [set(r) for r in routes]
# print(routes)
v_stop, v_bus = {source}, set()
q = deque([(source, 0)])
while q:
cur, res = q.popleft()
if cur == target: return res
for bus in stations[cur] - v_bus: # 在当前这个车站能坐哪些公交车
for stop in routes[bus] - v_stop: # 坐当前这个公交车能到哪些车站
v_bus.add(bus)
v_stop.add(stop)
q.append((stop, res + 1))
return -1
