Intelligent Robot

나의 풀이 n, m = map(int, input().split()) world = [] for _ in range(n): world.append(list(map(int,input()))) from collections import deque dr = [0, -1, 0, 1] dc = [1, 0, -1, 0] def bfs(r, c): q = deque() q.append([r,c]) while q: r, c = q.popleft() for i in range(4): nr = r + dr[i] nc = c + dc[i] if 0 = m: continue # 벽인 경우 무시 if graph[nx][ny] == 0: continue # 해당 노드를 처음 방문하는 경우에만 최단 거리 기록 if graph[nx..

나의 풀이 n, m = map(int, input().split()) world = [] for _ in range(n): world.append(list(map(int, input()))) def dfs(r, c): if r =n or c = m: return False else: if world[r][c] == 0: world[r][c] = 1 dfs(r-1, c) dfs(r+1, c) dfs(r, c-1) dfs(r, c+1) return True else: return False cnt = 0 for r in range(n): for c in range(m): if dfs(r, c): cnt += 1 print(cnt) 답안 예시 # N, M을 공백을 기준으로 ..

from collections import deque def bfs(graph, start, visited): queue = deque([start]) visited[start] = True while queue: v = queue.popleft() print(v, end = ' ') for i in graph[v]: if not visited[i]: queue.append(i) visited[i] = True graph = [ [], [2,3,8], [1,7], [1,4,5], [3,5], [3,4], [7], [2,6,8], [1,7] ] visited = [False] * 9 bfs(graph, 1, visited)

def dfs(graph, v, visited): visited[v] = True print(v, end=' ') for i in graph[v]: if not visited[i]: dfs(graph, i, visited) graph = [ [], [2,3,8], [1,7], [1,4,5], [3,5], [3,4], [7], [2,6,8], [1,7] ] visited = [False] * 9 dfs(graph, 1, visited)

def factorial_recursive(n): if n

from collections import deque queue = deque() queue.append(5) queue.append(2) queue.append(3) queue.append(7) queue.popleft() queue.append(1) queue.append(4) queue.popleft() print(queue) queue.reverse() print(queue)

stack = [] stack.append(5) stack.append(2) stack.append(3) stack.append(7) stack.pop() stack.append(1) stack.append(4) stack.pop() print(stack) print(stack[::-1])

나의 풀이 n, m = map(int, input().split()) a, b, d = map(int, input().split()) world = [] for _ in range(n): world.append(list(map(int, input().split()))) def turn_left(): global d if d == 0: d = 3 else: d -= 1 def visited(): global world world[a][b] = 2 cnt = 1 turn_cnt = 0 left_types = [[0,-1], [-1,0], [0,1], [1,0]] back_types = [[1,0], [0,-1], [-1,0], [0,1]] visited() while True: tmp_a, tmp_b = a..

나의 풀이 position = input() r = int(position[1]) c = ord(position[0]) - ord('a') + 1 move_types = [[2,1], [2,-1], [-2,1], [-2,-1], [1,2], [-1,2], [1,-2], [-1,-2]] cnt = 0 for m in move_types: tmp_r = r + m[0] tmp_c = c + m[1] if 0 < tmp_r < 9 and 0 < tmp_c < 9: cnt += 1 print(cnt) 답안 예시 # 현재 나이트의 위치 입력받기 input_data = input() row = int(input_data[1]) column = int(ord(input_data[0])) - int(ord('a')) ..

나의 풀이 n = int(input()) cnt = 0 for h in range(n+1): for m in range(60): for s in range(60): if '3' in str(h) + str(m) + str(s): cnt += 1 print(cnt) 답안 예시 # H를 입력받기 h = int(input()) count = 0 for i in range(h + 1): for j in range(60): for k in range(60): # 매 시각 안에 '3'이 포함되어 있다면 카운트 증가 if '3' in str(i) + str(j) + str(k): count += 1 print(count)