Baekjoon(백준)-7569 토마토(Gold 5)

baekjoon 7569 - 토마토문제입니다.

1. 문제

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

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2. Input , Output

baekjoon문제는 스크린샷을 포함하지 않습니다.

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3. 분류 및 난이도

Gold5

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4. 문제 해석

링크 참조.

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5. code

python

import sys
from collections import deque

tomato = []

M,N,H = map(int,input().split())

dq = deque()

for depth in range(H) : 
    tomato.append([])
    for col in range(N) :
        tomato[depth].append(list(map(int,sys.stdin.readline().split())))
            

# depth, col, row 순임.

check = [[[0 for row in range(M)]for col in range(N)] for depth in range(H)]



day = 0
tempdq = deque()

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

#초기 체크
tocount = 0 

for depth in range(H) :
    for row in range(M) : 
        for col in range(N) : 
            #print(depth,row,col,M,N)
            if tomato[depth][col][row] == 1: 
                check[depth][col][row] = 1
                dq.append((depth,row,col,0))
            if tomato[depth][col][row] == 0 :
                tocount += 1
#print(dq)

while dq : 
    d,r,c,counting = dq.popleft()
    #print(d,r,c,counting)
    for k in range(len(dz)) :
        newr,newc,newd = dx[k] + r,dy[k] + c,dz[k] + d
        if 0<= newd and newd < H and 0<=newr and newr<M and 0<=newc and newc < N :
            if check[newd][newc][newr] == 0 and tomato[newd][newc][newr] != -1 :
                if tomato[newd][newc][newr] == 0 :
                    tocount-=1
                tempdq.append((newd,newr,newc))
                dq.append((newd,newr,newc,counting+1))
                check[newd][newc][newr] = 1
    while tempdq : 
        td,tr,tc = tempdq.popleft()
        tomato[td][tc][tr] = 1
    day=counting

if tocount == 0 :
    print(day)
else : 
    print(-1)




                
            

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6. 결과 및 후기, 개선점

필요시 c++로 짜드립니다.

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