[백준] 3184번: 양 Python - BFS풀이

https://www.acmicpc.net/problem/3184

3184번: 양

전형적인 dfs/bfs 문제라고 생각함 

나는 bfs로 풀었다 

from collections import deque

r, c = map(int, input().split())
array = []
sheep = 0
wolf = 0
for i in range(r):
    array.append(input())

dx = [-1, 1, 0, 0]
dy = [0, 0, -1, 1]

visited = [[False] * c for _ in range(r)]


def bfs(a, b):
    global visited
    temp_sheep = 0
    temp_wolf = 0

    queue = deque([])
    queue.append((a, b))
    visited[a][b] = True
    if array[a][b] == 'o':
        temp_sheep += 1
    if array[a][b] == 'v':
        temp_wolf += 1
    while queue:
        x, y = queue.popleft()

        for i in range(4):
            nx = x + dx[i]
            ny = y + dy[i]

            if 0 <= nx < r and 0 <= ny < c and not visited[nx][ny] and array[nx][ny] != '#':
                if array[nx][ny] == 'o':
                    temp_sheep += 1
                    visited[nx][ny] = True
                    queue.append((nx, ny))
                if array[nx][ny] == 'v':
                    temp_wolf += 1
                    visited[nx][ny] = True
                    queue.append((nx, ny))
                if array[nx][ny] == '.': # .
                    visited[nx][ny] = True
                    queue.append((nx, ny))
    if temp_sheep > temp_wolf:
        return temp_sheep, 0
    else:
        return 0, temp_wolf


for i in range(r):
    for j in range(c):
        if array[i][j] != '#' and visited[i][j] == False:
                result = bfs(i, j)
                sheep += result[0]
                wolf += result[1]

print(sheep, wolf)