結果

問題 No.86 TVザッピング(2)
ユーザー mkawa2
提出日時 2023-03-06 23:37:19
言語 Python3
(3.13.1 + numpy 2.2.1 + scipy 1.14.1)
結果
AC  
実行時間 29 ms / 5,000 ms
コード長 1,945 bytes
コンパイル時間 247 ms
コンパイル使用メモリ 12,800 KB
実行使用メモリ 10,880 KB
最終ジャッジ日時 2024-09-18 01:57:02
合計ジャッジ時間 1,952 ms
ジャッジサーバーID
（参考情報） 		judge2 / judge3
このコードへのチャレンジ
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ファイルパターン 結果
sample AC * 3
other AC * 30
権限があれば一括ダウンロードができます

ソースコード

diff #
プレゼンテーションモードにする

import sys
sys.setrecursionlimit(200005)
int1 = lambda x: int(x)-1
pDB = lambda *x: print(*x, end="\n", file=sys.stderr)
p2D = lambda x: print(*x, sep="\n", end="\n\n", file=sys.stderr)
def II(): return int(sys.stdin.readline())
def LI(): return list(map(int, sys.stdin.readline().split()))
def LLI(rows_number): return [LI() for _ in range(rows_number)]
def LI1(): return list(map(int1, sys.stdin.readline().split()))
def LLI1(rows_number): return [LI1() for _ in range(rows_number)]
def SI(): return sys.stdin.readline().rstrip()
dij = [(0, 1), (-1, 0), (0, -1), (1, 0)]
# dij = [(0, 1), (-1, 0), (0, -1), (1, 0), (1, 1), (1, -1), (-1, 1), (-1, -1)]
inf = (1 << 63)-1
# inf = (1 << 31)-1
md = 10**9+7
# md = 998244353
def start():
    for i in range(h):
        for j, a in enumerate(aa[i]):
            if aa[i][j] == 0:
                return i, j
def dfs(i, j, d, r):
    if i == si and j == sj and aa[i][j]:
        for i in range(h):
            for a in aa[i]:
                if a == 0: return False
        if r and (d-1)%4 == d0: return False
        return True
    di, dj = dij[d]
    ni, nj = i+di, j+dj
    if 0 <= ni < h and 0 <= nj < w and aa[ni][nj] == 0:
        aa[ni][nj] = 1
        if dfs(ni, nj, d, r): return True
        aa[ni][nj] = 0
    else:
        d = (d+1)%4
        di, dj = dij[d]
        ni, nj = i+di, j+dj
        if 0 <= ni < h and 0 <= nj < w and aa[ni][nj] == 0:
            aa[ni][nj] = 1
            if dfs(ni, nj, d, r): return True
            aa[ni][nj] = 0
        d = (d+2)%4
        di, dj = dij[d]
        ni, nj = i+di, j+dj
        if 0 <= ni < h and 0 <= nj < w and aa[ni][nj] == 0 and r == 0:
            aa[ni][nj] = 1
            if dfs(ni, nj, d, 1): return True
            aa[ni][nj] = 0
    return False
h, w = LI()
aa = [[c == "#" for c in SI()] for _ in range(h)]
si, sj = start()
for d0 in range(4):
    if dfs(si, sj, d0, 0):
        print("YES")
        exit()
print("NO")
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