网站开发和网络开发区别seo店铺描述

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：

3
A B C D F
C E G H 
B G I J
A J

输出：

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

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 = Falsefor 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] * Nprev = {}q = deque()for line in start_lines:dist[line] = 0q.append(line)target_line = -1found = Falsewhile q:current_line = q.popleft()if current_line in end_lines:target_line = current_linefound = Truebreakfor 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] + 1prev[neighbor_line] = (current_line, s)q.append(neighbor_line)if not found:print("NA")exit()else:path = []current_line = target_linewhile current_line in prev:current_line, s = prev[current_line]path.append(s)path.reverse()res_path = startfor i in range(len(path)):res_path = res_path + '-' + path[i]res_path = res_path + '-' + endprint(res_path)k = len(path)cost = k + 2print(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()