結果
問題 | No.1295 木と駒 |
ユーザー |
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提出日時 | 2020-11-23 16:28:44 |
言語 | PyPy3 (7.3.15) |
結果 |
AC
|
実行時間 | 402 ms / 2,000 ms |
コード長 | 1,361 bytes |
コンパイル時間 | 176 ms |
コンパイル使用メモリ | 82,560 KB |
実行使用メモリ | 114,508 KB |
最終ジャッジ日時 | 2024-07-23 17:34:32 |
合計ジャッジ時間 | 13,314 ms |
ジャッジサーバーID (参考情報) |
judge3 / judge5 |
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ファイルパターン | 結果 |
---|---|
sample | AC * 1 |
other | AC * 48 |
ソースコード
import sysinput = sys.stdin.readlinefrom collections import dequeN = int(input())X = [[] for i in range(N)]for i in range(N-1):x, y = map(int, input().split())x, y = x-1, y-1X[x].append(y)X[y].append(x)MI = [min(x) for x in X]MA = [max(x) for x in X]MA2 = [max([y for y in x + [-1] if y != ma]) for x, ma in zip(X, MA)]P = [-1] * NQ = deque([0])R = []while Q:i = deque.popleft(Q)R.append(i)for a in X[i]:if a != P[i]:P[a] = iX[a].remove(i)deque.append(Q, a)##### Settingsunit = 0merge = lambda a, b: a + bdef adj_bu(a, i, p):if a >= 4 or (a == 3 and p != MA[i]): return 4if a == 0 and p == MI[i]: return 0if i == MA[p] or i == MI[p]: return 2if i == MA2[p]: return 3return 4adj_td = lambda a, i, p: adj_bu(a, p, i)adj_fin = lambda a, i: "Yes" if a <= 2 else "No"#####ME = [unit] * NXX = [0] * NTD = [unit] * Nfor i in R[1:][::-1]:p = P[i]XX[i] = adj_bu(ME[i], i, p)ME[p] = merge(ME[p], XX[i])XX[R[0]] = adj_fin(ME[R[0]], R[0])for i in R:ac = TD[i]for j in X[i]:TD[j] = acac = merge(ac, XX[j])ac = unitfor j in X[i][::-1]:TD[j] = adj_td(merge(TD[j], ac), j, i)ac = merge(ac, XX[j])XX[j] = adj_fin(merge(ME[j], TD[j]), j)print("\n".join(XX))