프로그래머스 dfs bfs 여행경로

문제

 

잘못한 풀이

from collections import deque
#1. 같은 항공권이 중복될 수 있음
#2. 모든 도시를 방문할 수 없는 알고리즘?
def solution(tickets):
    answer = []
    graph = []
    visited = []
    all_dict = {}
    all_dict_revers = {}
    
    #print("a"<"b")
    
    def make_graph():
        nonlocal graph, all_dict, visited, all_dict_revers
        
        all = []
        for start, end in tickets:
            all.append(start)
            all.append(end)

        all = set(all)
        all = list(all)
        all = sorted(all)

        for idx, city in enumerate(all):
            all_dict[city] = idx
        
        for city, idx in all_dict.items():
            all_dict_revers[idx] = city
            
        all_len = len(all)

        graph = [[0]*(all_len) for _ in range(all_len)]
        visited =  [[0]*(all_len) for _ in range(all_len)]
        for start, end in tickets:
            start_num = all_dict[start]
            end_num = all_dict[end]
            graph[start_num][end_num] = 1

        
    def bfs():
        nonlocal answer
        q = deque()
        q.append("ICN")
        while(len(q)>=1):
            cur_city = q.popleft()
            cur_num = all_dict[cur_city]
            answer.append(cur_city)
            
            for next_num in range(len(graph[cur_num])):
                
                if graph[cur_num][next_num]==0:
                    continue
                if visited[cur_num][next_num]==1:
                    continue
                next_city = all_dict_revers[next_num]
                visited[cur_num][next_num] = 1
                q.append(next_city)
                break
        
        
    
    make_graph()
    bfs()
    # for start, end in tickets:   
    #print('graph',graph, 'all_dict',all_dict, "all_dict reverse", all_dict_revers)
    
    return answer