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Binary Journey
[이코테] dfs & bfs 문제풀이 본문
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graph = []
n, m = map(int, "4 5".split())
example = [
"00110",
"00011",
"11111",
"00000"
]
for i in example:
graph.append(list(map(int, str(i))))
def dfs(i, j):
if i <= -1 or i >= n or j <= -1 or j >= m:
return False
if graph[i][j] == 0:
graph[i][j] = 1
dfs(i - 1, j)
dfs(i + 1, j)
dfs(i, j - 1)
dfs(i, j + 1)
return True
return False
result = 0
for i in range(n):
for j in range(m):
if dfs(i, j):
result += 1
print(result)
graph = []
n, m = map(int, "15 14".split())
example = [
"00000111100000",
"11111101111110",
"11011101101110",
"11011101100000",
"11011111111111",
"11011111111100",
"11000000011111",
"01111111111111",
"00000000011111",
"01111111111000",
"00011111111000",
"00000001111000",
"11111111110011",
"11100011111111",
"11100011111111"
]
for i in example:
graph.append(list(map(int, str(i))))
def dfs(i, j):
if i <= -1 or i >= n or j <= -1 or j >= m:
return False
if graph[i][j] == 0:
graph[i][j] = 1
dfs(i - 1, j)
dfs(i + 1, j)
dfs(i, j - 1)
dfs(i, j + 1)
return True
return False
result = 0
for i in range(n):
for j in range(m):
if dfs(i, j):
result += 1
print(result)
from collections import deque
def bfs(n, m, graph, x, y):
queue = deque()
queue.append((x, y))
while queue:
a, b = queue.popleft()
for i in range(4):
nx = a + dx[i]
ny = b + dy[i]
if nx < 0 or nx >= n or ny < 0 or ny >= m:
continue
if graph[nx][ny] == 0:
continue
if graph[nx][ny] == 1:
graph[nx][ny] = graph[a][b] + 1
queue.append((nx, ny))
return graph[n - 1][m - 1]
n, m = 5, 6
graph = [
[1, 0, 1, 0, 1, 0],
[1, 1, 1, 1, 1, 1],
[0, 0, 0, 0, 0, 1],
[1, 1, 1, 1, 1, 1],
[1, 1, 1, 1, 1, 1]
]
dx = [-1, 1, 0, 0]
dy = [0, 0, -1, 1]
x = 0
y = 0
print(bfs(n, m, graph, x, y))
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