[백준] #7569 Python 토마토의 토는 오바이트의 토

https://www.acmicpc.net/problem/7569

 

시간초과 뜬 내 코드

import sys
from collections import deque

input = sys.stdin.readline

m, n, h = map(int, input().split())

graph = []
for _ in range(n * h):
    graph.append(list(map(int, input().split())))

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

def bfs(i, j):
    q = deque()
    q.append((i, j))
    
    while q:
        tempx, tempy = q.popleft()
        for idx in range(6):
            x = tempx + dx[idx]
            y = tempy + dy[idx]
            if 0 <= x < n * h and 0 <= y < m and ((graph[x][y] >= 1 and graph[x][y] > graph[tempx][tempy] + 1) or graph[x][y] == 0):
                q.append((x, y))
                graph[x][y] = graph[tempx][tempy] + 1

#이중배열에서 원소 찾기
if all(0 not in l for l in graph):
    print(0)
else:
    for i in range(n * h):
        for j in range(m):
            if graph[i][j] == 1:
                bfs(i, j)
    if any(0 in ele for ele in graph):
        print(-1)
    else:
        maxval = 0
        for el in graph:
            maxval = max(maxval, max(el))
        print(maxval - 1)

이 코드 짜면서 출력할 값을 분화했는데

if all(0 not in l for l in graph):

if any(0 in i for i in graph):

이런 표현 찾았는데

이렇게 짜지 않아도 더 효율적으로 짤 수 있었다..!

시간 초과도.. bfs 함수를 for 문 안에 넣으면서 더 돌아갔나..? 

멋쟁이 스터디원!!!

 

스터디원 코드

from sys import stdin
from collections import deque

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

queue = deque([])

def bfs():
	while queue:
		x, y, z = queue.popleft()
		for i in range(6):
			nx = x + dx[i]
			ny = y + dy[i]
			nz = z + dz[i]
			if 0 <= nx < h and 0 <= ny < n and 0 <= nz < m and not graph[nx][ny][nz]:
				graph[nx][ny][nz] = graph[x][y][z] + 1
				queue.append([nx, ny, nz])

m, n, h = map(int, stdin.readline().split())
graph = [[list(map(int, stdin.readline().split())) for _ in range(n)] for _ in range(h)]

for i in range(h):
	for j in range(n):
		for k in range(m):
			if graph[i][j][k] == 1:
				queue.append([i, j, k])
bfs()

answer = 0
for i in graph:
	for j in i:
		if 0 in j:
			print(-1)
			exit(0)
		answer = max(answer, max(j))
print(answer - 1)