TA Engineer

■ 문제풀이 1) dfs 알고리즘 ■ 해답 n = int(input()) graph = [] for i in range(n): graph.append(list(map(int, input()))) def dfs(x, y): global count if x = n or y = n: return False if graph[x][y] == 1: count += 1 graph[x][y] = 0 dfs(x - 1, y) dfs(x + 1, y) dfs(x, y - 1) dfs(x, y + 1) return True return False result = 0 danji = [] count = 0 for i in range(n): for j in range(n): if dfs(i, j) == True: result +..

■ 문제풀이 1) dfs 알고리즘 ■ 해답 n = int(input()) m = int(input()) graph = [[] for _ in range(n + 1)] for i in range(m): a, b = map(int, input().split()) graph[a].append(b) graph[b].append(a) graph[a].sort() graph[b].sort() visited = [False] * (n + 1) cnt = 0 def dfs(graph, v, visited): global cnt visited[v] = True for i in graph[v]: if not visited[i]: cnt += 1 dfs(graph, i, visited) dfs(graph, 1, visite..

■ 해답 n, m, v = map(int, input().split()) graph = [[] for _ in range(n + 1)] for _ in range(m): a, b = map(int, input().split()) graph[a].append(b) graph[b].append(a) graph[a].sort() graph[b].sort() visited_dfs = [False] * (n + 1) visited_bfs = [False] * (n + 1) def dfs(graph, v, visited_dfs): visited_dfs[v] = True print(v, end = ' ') for i in graph[v]: if not visited_dfs[i]: dfs(graph, i, visite..