DFS10 [python] 17472. 다리 만들기 2 문제 출처 17472. 다리 만들기 2 풀이 from collections import deque N, M = map(int, input().split()) arr = [list(map(int, input().split())) for _ in range(N)] visited = [[0]*M for _ in range(N)] dx, dy = (1, 0, -1, 0), (0, 1, 0, -1) def numbering_irlands(x, y): visited[y][x] = 1 arr[y][x] = irland q = deque() q.append((x, y)) while q: x, y = q.popleft() for i in range(4): nx, ny = x + dx[i], y + dy[i] if nx .. 2019. 10. 19. [python] 2146. 다리 만들기 문제 출처 2146. 다리 만들기 풀이 from collections import deque # 섬에 넘버링 def numbering_irlands(x, y): arr[y][x] = irland q = deque() q.append((x, y)) while q: x, y = q.popleft() visited[y][x] = 1 for i in range(4): nx, ny = x + dx[i], y + dy[i] if nx = N or ny = N: continue if not visited[ny][nx] and arr[ny][nx]: arr[ny][nx] = irland q.append((nx, ny)) visited[ny][nx] = 1 def get_min_d.. 2019. 10. 17. [python] 단어 변환 문제 출처 단어 변환 풀이 def dfs(now, target, words, history=[], depth=0): global min_ print(now, history) if now == target: min_ = min(min_, depth) if depth >= min_: return for word in words: if word in history: continue cnt = 0 for i in range(len(target)): if word[i] != now[i]: cnt += 1 if cnt > 1: break if cnt == 1 and word not in history: history.append(word) dfs(word, target, words, history, depth+1).. 2019. 9. 1. [python] 네트워크 문제 출처 네트워크 풀이 def solution(n, computers): answer = 0 n = len(computers) while 1: linked = [] for i in range(n): if sum(computers[i]) > 1: for j in range(n): if computers[i][j]: computers[i][j] = 0 computers[j][i] = 0 if i != j: linked.append((i, j)) break if not linked: for y in range(n): if sum(computers[y]): answer += 1 break answer += 1 while linked: y, x = linked.pop(0) for i in range(n): if.. 2019. 9. 1. 이전 1 2 3 다음
