기존에는 dfs와 bfs 연산을 나누며, 방문한 노드들을 저장한 배열에서 확인하는 연산 방법에 확인하는 방법으로 풀었다. 이번 풀이는 한 함수에서 dfs와 bfs 연산을 구분하여 연산하며 방문확인 배열을 만들어 방문확인에 소요되는 연산을 줄였다.
\n# from collections import deque\n#\n# def dfs(gra, sta):\n# vis = []\n# stack = deque([sta])\n# while stack:\n# nod = stack.pop()\n# if nod not in vis:\n# vis.append(nod)\n# if nod in gra:\n# stack.extend(sorted(gra[nod], reverse=True))\n# return \" \".join(str(i) for i in vis)\n#\n# def bfs(gra, sta):\n# vis = []\n# que = deque([sta])\n# while que:\n# nod = que.popleft()\n# if nod not in vis:\n# vis.append(nod)\n# if nod in gra:\n# que.extend(sorted(gra[nod]))\n#\n# return \" \".join(str(i) for i in vis)\n#\n# graph = {}\n# n, m, start = map(int, input().split())\n# for _ in range(m):\n# x, y = map(int, input().split())\n# if x in graph:\n# graph[x].append(y)\n# else:\n# graph[x] = [y]\n# if y in graph:\n# graph[y].append(x)\n# else:\n# graph[y] = [x]\n#\n# print(dfs(graph, start))\n# print(bfs(graph, start))\n\nimport sys\nfrom collections import deque\ninput = sys.stdin.readline\n\ndef solve(start, is_bfs):\n q = deque() if is_bfs else []\n q.append(start)\n visit = [False for _ in range(n+1)]\n ans = []\n\n while q:\n now = q.popleft() if is_bfs else q.pop()\n if not visit[now]:\n visit[now] = True\n ans.append(str(now))\n if now in graph:\n q.extend(sorted(graph[now]) if is_bfs else sorted(graph[now], reverse=True))\n\n return \" \".join(ans)\n\nn, m, v = map(int, input().split())\ngraph = {}\nfor _ in range(m):\n s, e = map(int, input().split())\n if s in graph: graph[s].append(e)\n else: graph[s] = [e]\n if e in graph: graph[e].append(s)\n else: graph[e] = [s]\n\nprint(solve(v, False))\nprint(solve(v, True))