結果

問題 No.2984 [Cherry Anniversary 4] 満開の願いを込めた 27 の桜
ユーザー mkawa2
提出日時 2024-12-09 12:50:30
言語 PyPy3
(7.3.15)
結果
AC  
実行時間 284 ms / 4,000 ms
コード長 2,214 bytes
コンパイル時間 322 ms
コンパイル使用メモリ 82,408 KB
実行使用メモリ 80,024 KB
最終ジャッジ日時 2024-12-09 12:50:37
合計ジャッジ時間 6,491 ms
ジャッジサーバーID
（参考情報） 		judge4 / judge5
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ファイルパターン 結果
other AC * 22
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)
# sys.set_int_max_str_digits(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-(-1 << 31)
inf = -1-(-1 << 62)
# md = 10**9+7
md = 998244353
from itertools import permutations
from collections import deque, defaultdict
pp = list(permutations(range(3)))
pn = len(pp)
cost = defaultdict(lambda: -1)
cost[pp[0]] = 0
q = deque()
q.append(pp[0])
while q:
    p = q.popleft()
    c = cost[p]
    for i in range(3):
        for j in range(3):
            l = list(p)
            l[i], l[j] = l[j], l[i]
            np = tuple(l)
            if cost[np] == -1:
                cost[np] = c+1
                q.append(np)
def solve():
    def ok():
        for i in range(3):
            ni = p[i]
            for j in range(3):
                nj = q[j]
                if zz[i][j] != ij[ni][nj]: return False
                for k in range(3):
                    nk = r[k]
                    if xx[j][k] != jk[nj][nk]: return False
                    if yy[i][k] != ik[ni][nk]: return False
        return True
    aa = [LLI(3) for _ in range(3)]
    xx = LLI(3)
    yy = LLI(3)
    zz = LLI(3)
    jk = [[0]*3, [0]*3, [0]*3]
    ik = [[0]*3, [0]*3, [0]*3]
    ij = [[0]*3, [0]*3, [0]*3]
    for i in range(3):
        for j in range(3):
            for k, a in enumerate(aa[i][j]):
                jk[j][k] += a
                ik[i][k] += a
                ij[i][j] += a
    ans = inf
    for p in pp:
        for q in pp:
            for r in pp:
                if ok(): ans = min(ans, cost[p]+cost[q]+cost[r])
    if ans == inf: ans = -1
    print(ans)
for _ in range(II()): solve()
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