結果
問題 |
No.3199 Key-Door Grid
|
ユーザー |
![]() |
提出日時 | 2025-07-24 12:25:27 |
言語 | PyPy3 (7.3.15) |
結果 |
AC
|
実行時間 | 406 ms / 3,000 ms |
コード長 | 3,954 bytes |
コンパイル時間 | 575 ms |
コンパイル使用メモリ | 82,220 KB |
実行使用メモリ | 109,760 KB |
最終ジャッジ日時 | 2025-07-24 12:25:36 |
合計ジャッジ時間 | 8,545 ms |
ジャッジサーバーID (参考情報) |
judge5 / judge2 |
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ファイルパターン | 結果 |
---|---|
sample | AC * 3 |
other | AC * 37 |
ソースコード
#yukicoder contest 473 第1回 生成AI作問コンテスト ''' #A S = input().split() print(''.join(Si[0].upper() for Si in S)) #B N = int(input()) names = [input().split() for _ in range(N)] #この問題AtCoderで見たな・・・? 名前を座標圧縮 S = [] for Si, Ti in names: S.append(Si) S.append(Ti) back = None S = [back := Si for Si in sorted(S) if back != Si] import bisect names = [(bisect.bisect_left(S, Si), bisect.bisect_left(S, Ti)) for Si, Ti in names] n = len(S) C = [0] * n for Si, Ti in names: C[Si] += 1 C[Ti] += 1 print('Yes' if all(C[Si] == 1 or C[Ti] == 1 for Si, Ti in names) else 'No') #C import bisect N = int(input()) A = list(map(int, input().split())) back = None B = [back := Ai for Ai in sorted(A) if back != Ai] C = [0] * ( n := len(B) ) for Ai in A: C[ bisect.bisect_left(B, Ai) ] += 1 for _ in range( Q := int(input()) ): Xi, Ki = map(int, input().split()) print('Yes' if 0 <= ( i := bisect.bisect_left(B, Xi) ) < n and B[i] == Xi and C[i] >= Ki else 'No') #D #どうしよっかな SegTree使っていい? #Segment Tree class SegmentTree: def __init__(self, N, identity_e, node_f): self._N, self._size, self._e, self._f = N, 1 << N.bit_length(), identity_e, node_f self._node = [self._e for _ in range(self._size << 1)] def build(self, A): assert len(A) == self._N, 'array too large' for i, v in enumerate(A, start = self._size): self._node[i] = v for i in range(self._size - 1, 0, -1): self._node[i] = self._f( self._node[i << 1], self._node[i << 1 | 1] ) def update(self, i, v): self._node[i := i + self._size] = v while (i := i >> 1) != 0: self._node[i] = self._f( self._node[i << 1], self._node[i << 1 | 1] ) def fold(self, Lt, Rt): if not 0 <= Lt < Rt <= self._N: return self._e Lt, Rt, vL, vR = Lt | self._size, Rt + self._size, self._e, self._e while Lt < Rt: if Lt & 1: vL = self._f( vL, self._node[Lt] ); Lt += 1 if Rt & 1: Rt -= 1; vR = self._f( self._node[Rt], vR ) Lt >>= 1; Rt >>= 1 return self._f( vL, vR ) ST = SegmentTree( Q := int(input()), 0, max ) Rt = 0 for _ in range(Q): Ti, Xi = map(int, input().split()) match Ti: case 1: ST.update(Rt, Xi) Rt += 1 case 2: Lt = Rt - Xi print(ST.fold(Lt, Rt)) ''' #E #複雑だな 鍵を持っていない状態を0としたいから・・・ #通路を0 鍵を1 ~ 9 扉を11 ~ 19 壁を20 に対応させればいいかな H, W, M = map(int, input().split()) l = W.bit_length() B = bytearray(H + 1 << l) for i in range( len(B) ): B[i] = 20 for h in range(H): for w, Shw in enumerate(input().rstrip()): i = h << l | w if Shw == '.': B[i] = 0 elif Shw == '#': B[i] = 20 elif Shw == 'S': B[i] = 0; st = i elif Shw == 'G': B[i] = 0; gl = i elif Shw in '123456789': B[i] = int(Shw) elif Shw in 'abcdefghi': B[i] = 11 + ord(Shw) - ord('a') else: assert False #DP[i][f]: 現在値がi、鍵がfとなるような最短経路 m = M.bit_length() DP = [10 ** 9] * (len(B) << m) DP[st << m | 0] = 0 maskm = ~ ( - 1 << m ) Q, R = [], [st << m | 0] dist = -1 while R: Q, R = R, Q dist += 1 while Q: x = Q.pop() if DP[x] < dist: continue DP[x] = dist i, f = x >> m, x & maskm for t in -1, 1, 1 << l, -1 << l: if B[j := i + t] == 20: continue if B[j] == 0: g = f elif 1 <= B[j] <= 9: g = B[j] elif 11 <= B[j] <= 19: if f != B[j] - 10: continue else: g = f if DP[j << m | g] > dist + 1: DP[j << m | g] = dist + 1 R.append(j << m | g) ans = min(DP[gl << m | f] for f in range(M + 1)) print(ans if ans != 10 ** 9 else -1)