题目
已知某市运营了N条地铁线路，市民在乘坐地铁时单条线路通票2元，换乘一次加1元。给出N条线路的所有站名列表，请帮乘客寻找从出发站到目的站最便宜的地铁换乘方案，并输出票价。每条地铁线路不包含环路，即没有相同站名。
输入：
第一行为地铁线路条数N，范围是[1, 1000]
第二行到第N+1行，每条线路依次包含的站名，每个站名包含的字符个数不超过100，站点个数也不超过100，依次用空格隔开，不同线路中相同的站点名表示是一个换乘站
第N+2行，出发站和目的站，用空格隔开
输入保证：若可达则为唯一解
输出：
第一行按出发站-换乘站（可以是多个）-目的站的格式输出换乘方案的字符串
第二行输出换乘方案的总票价
如果没有任何方案实现出发站到目的站，只输出一行：NA
样例：
输入：
python
3
A B C D F
C E G H
B G I J
A J输出：
python
A-B-J
3解释：一号线有A,B,C,D,F五个站点，二号线有C,E,G,H四个站点，三号线有B,G,I,J四个站点。从A站到J站的最省钱换乘路线是经B站换乘，票价3元
代码
思路：输入处理------站点与线路建立映射------直接到达检查------邻接表构建------BFS------路径回溯与打印输出
python
from collections import defaultdict, deque
# If you need to import additional packages or classes, please import here.
def func():
N = int(input())
lines = []
for i in range(N):
line = list(map(str, input().split()))
lines.append(line)
start, end = list(map(str, input().split()))
station_to_lines = defaultdict(list)
for line_idx, stations in enumerate(lines):
for s in stations:
station_to_lines[s].append(line_idx)
found_direct = False
for line_idx in station_to_lines[start]:
if end in lines[line_idx]:
print(f"{start}-{end}")
print(2)
exit()
adj_with_s = defaultdict(list)
for s in station_to_lines:
line_list = station_to_lines[s]
for i in range(len(line_list)):
for j in range(i + 1, len(line_list)):
l1 = line_list[i]
l2 = line_list[j]
adj_with_s[l1].append((l2, s))
adj_with_s[l2].append((l1, s))
start_lines = station_to_lines[start]
end_lines = set(station_to_lines[end])
dist = [-1] * N
prev = {}
q = deque()
for line in start_lines:
dist[line] = 0
q.append(line)
target_line = -1
found = False
while q:
current_line = q.popleft()
if current_line in end_lines:
target_line = current_line
found = True
break
for neighbor_line, s in adj_with_s.get(current_line, []):
if dist[neighbor_line] == -1 or dist[current_line] + 1 < dist[neighbor_line]:
dist[neighbor_line] = dist[current_line] + 1
prev[neighbor_line] = (current_line, s)
q.append(neighbor_line)
if not found:
print("NA")
exit()
else:
path = []
current_line = target_line
while current_line in prev:
current_line, s = prev[current_line]
path.append(s)
path.reverse()
res_path = start
for i in range(len(path)):
res_path = res_path + '-' + path[i]
res_path = res_path + '-' + end
print(res_path)
k = len(path)
cost = k + 2
print(cost)
# please define the python3 input here. For example: a,b = map(int, input().strip().split())
# please finish the function body here.
# please define the python3 output here. For example: print().
if __name__ == "__main__":
func()