n, m = map(int, input().split())
graph = []
for i in range(n):
graph.append(list(map(int, input())))
def dfs(x, y):
if x <= -1 or x >= n or y <= -1 or y >= m:
return False
if graph[x][y] == 0:
graph[x][y] = 1
# 상, 하, 좌, 우의 위치도 모두 재귀적으로 호출
dfs(x - 1, y)
dfs(x, y - 1)
dfs(x + 1, y)
dfs(x, y + 1)
return True
return False
result = 0
for i in range(n):
for j in range(m):
if dfs(i, j) == True:
result += 1
print(result)
def solution(numbers, target):
n = len(numbers)
answer = 0
def dfs(idx, result):
if idx == n:
if result == target:
nonlocal answer
answer += 1
return
else:
dfs(idx+1, result+numbers[idx])
dfs(idx+1, result-numbers[idx])
dfs(0,0)
return answer
간선 하나로 갈 때의 경우만 보였으므로, DFS 사용
def solution(n, computers):
def DFS(i):
visited[i] = 1
for a in range(n):
if computers[i][a] and not visited[a]:
DFS(a)
answer = 0
visited = [0 for i in range(len(computers))]
for i in range(n):
if not visited[i]:
DFS(i)
answer += 1
return answer
from collections import deque
n, m = map(int, input().split())
graph = []
for i in range(n):
graph.append(list(map(int, input())))
dx = [-1, 1, 0, 0]
dy = [0, 0, -1, 1]
def bfs(x, y):
queue = deque()
queue.append((x, y)
while queue:
x, y = queue.popleft()
# 현재 위치에서 네 방향으로의 위치 확인
for i in range(4):
nx = x + dx[i]
ny = x + dy[i]
# 미로 찾기 공간을 벗어난 경우 무시
if nx < 0 or ny < 0 or nx >= n or ny >= m:
continue
# 해당 노드를 처음 방문하는 경우에만 최단 거리 기록
if graph[nx][ny] == 1:
graph[nx][ny] = graph[x][y] + 1
queue.append((nx, ny))
# 가장 오른쪽 아래까지의 최단 거리 반환
return graph[n - 1][m - 1]
print(bfs(0, 0))
#BFS 풀이
def solution(numbers, target):
answer = 0
leaves = [0]
for num in numbers:
tmp = []
for parent in leaves:
tmp.append(parent + num)
tmp.append(parent - num)
leaves = tmp
for leaf in leaves:
if leaf == target:
answer += 1
return answer
from collections import deque
def solution(n, computers):
def BFS(i):
q = deque()
q.append(i)
while q:
i = q.popleft()
visited[i] = 1
for a in range(n):
if computers[i][a] and not visited[a]:
q.append(a)
answer = 0
visited = [0 for i in range(len(computers))]
for i in range(n):
if not visited[i]:
BFS(i)
answer += 1
return answer