BOJ 17406 - 배열 돌리기 4

문제

백준 온라인 저지 - 17406번

풀이 과정

주어진 조건대로 배열 A 를 돌리며 배열의 최소값을 찾는 브루트포스 문제입니다.

회전 연산의 개수 K 가 크지 않기 때문에 순열을 생성하고 모든 경우에 대응해서 최소값을 찾으면 됩니다.

코드

import sys
import copy
from itertools import permutations

N, M, K = list(map(int, sys.stdin.readline().split()))
board = [list(map(int, sys.stdin.readline().split())) for _ in range(N)]
operation = [list(map(int, sys.stdin.readline().split())) for _ in range(K)]


def rotate(board, sx, sy, offset):
    while offset > 0:
        left_top_pos = [sx - offset, sy - offset]
        right_bottom_pos = [sx + offset, sy + offset]

        temp = board[left_top_pos[0] - 1][left_top_pos[1] - 1]

        for x in range(left_top_pos[0], right_bottom_pos[0]):
            board[x - 1][left_top_pos[1] - 1] = board[x][left_top_pos[1] - 1]

        for y in range(left_top_pos[1], right_bottom_pos[1]):
            board[right_bottom_pos[0] - 1][y - 1] = board[right_bottom_pos[0] - 1][y]

        for x in range(right_bottom_pos[0], left_top_pos[0], -1):
            board[x - 1][right_bottom_pos[1] - 1] = board[x - 2][right_bottom_pos[1] - 1]

        for y in range(right_bottom_pos[1], left_top_pos[1], -1):
            board[left_top_pos[0] - 1][y - 1] = board[left_top_pos[0] - 1][y - 2]

        board[left_top_pos[0] - 1][left_top_pos[1]] = temp

        offset -= 1


def get_array_value(board):
    ret = sys.maxsize
    for row in board:
        ret = min(ret, sum(row))
    return ret


def solution():
    answer = sys.maxsize

    for perms in permutations(operation):
        rotated_board = copy.deepcopy(board)
        for perm in perms:
            rotate(rotated_board, *perm)
        answer = min(answer, get_array_value(rotated_board))

    return answer


print(solution())