7.[再]トリクロラミノ
24809 ワード
白駿
1.シミュレーション
import sys
input = sys.stdin.readline
n, m = map(int, input().split())
board = [list(map(int, input().split())) for _ in range(n)]
tetromino = [
[(0, 0), (0, 1), (0, 2), (0, 3)], # 1
[(0, 0), (1, 0), (2, 0), (3, 0)],
[(0, 0), (0, 1), (1, 0), (1, 1)], # 2
[(0, 0), (0, 1), (1, 0), (2, 0)], # 3
[(0, 0), (0, 1), (1, 1), (2, 1)],
[(0, 0), (1, 0), (1, 1), (1, 2)],
[(1, 0), (1, 1), (1, 2), (0, 2)],
[(2, 0), (2, 1), (1, 1), (0, 1)],
[(0, 0), (1, 0), (2, 0), (2, 1)],
[(0, 0), (0, 1), (0, 2), (1, 2)],
[(0, 0), (0, 1), (0, 2), (1, 0)], # 4
[(1, 0), (2, 0), (0, 1), (1, 1)],
[(0, 0), (1, 0), (1, 1), (2, 1)],
[(0, 0), (0, 1), (1, 1), (1, 2)],
[(1, 0), (0, 1), (1, 1), (0, 2)],
[(1, 0), (1, 1), (1, 2), (0, 1)], #5
[(1, 0), (0, 1), (1, 1), (2, 1)],
[(0, 0), (1, 0), (2, 0), (1, 1)],
[(0, 0), (0, 1), (0, 2), (1, 1)],
]
answer = 0
def find(x, y):
global answer
for i in range(19):
result = 0
for j in range(4):
nx = x + tetromino[i][j][0]
ny = y + tetromino[i][j][1]
if 0 <= nx < n and 0 <= ny < m:
result += board[nx][ny]
answer = max(answer, result)
for i in range(n):
for j in range(m):
find(i, j)
print(answer)
Reference
この問題について(7.[再]トリクロラミノ), 我々は、より多くの情報をここで見つけました https://velog.io/@corone_hi/7.-re-테트로미노テキストは自由に共有またはコピーできます。ただし、このドキュメントのURLは参考URLとして残しておいてください。
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