結果
問題 |
No.1424 Ultrapalindrome
|
ユーザー |
|
提出日時 | 2021-03-13 12:15:54 |
言語 | PyPy3 (7.3.15) |
結果 |
TLE
|
実行時間 | - |
コード長 | 2,060 bytes |
コンパイル時間 | 329 ms |
コンパイル使用メモリ | 82,432 KB |
実行使用メモリ | 97,664 KB |
最終ジャッジ日時 | 2024-10-15 04:52:23 |
合計ジャッジ時間 | 6,173 ms |
ジャッジサーバーID (参考情報) |
judge4 / judge2 |
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ファイルパターン | 結果 |
---|---|
sample | AC * 3 |
other | AC * 15 TLE * 1 -- * 13 |
ソースコード
from collections import deque class LCA: "0-indexed" __slots__ = ["depth", "ancestor"] def __init__(self, adj): N = len(adj) parent = [-1] * N self.depth = [0] * N q = deque([0]) while q: node = q.popleft() for next_node in adj[node]: if parent[node] != next_node: parent[next_node] = node q.append(next_node) self.depth[next_node] = self.depth[node] + 1 self.ancestor = [parent] #self.ancestor[k][u]はuの2**k先の祖先。 k = 1 while (1 << k) < N: anc_k = [0] * N for u in range(N): if self.ancestor[-1][u] == -1: anc_k[u] = -1 else: anc_k[u] = self.ancestor[-1][self.ancestor[-1][u]] self.ancestor.append(anc_k) k += 1 def lca(self, u, v): if self.depth[u] < self.depth[v]: u, v = v, u for k, bit in enumerate(reversed(format(self.depth[u]-self.depth[v], 'b'))): if bit == '1': u = self.ancestor[k][u] if u == v: return u for anc in reversed(self.ancestor): if anc[u] != anc[v]: u = anc[u] v = anc[v] return self.ancestor[0][u] def dist(self, u, v): w = self.lca(u, v) return self.depth[u] + self.depth[v] - 2 * self.depth[w] def main(): N = int(input()) adj = [[] for _ in range(N)] for _ in range(N - 1): u, v = map(int, input().split()) u -= 1; v -= 1 adj[u].append(v) adj[v].append(u) lca = LCA(adj) leaves = [v for v in range(N) if len(adj[v]) == 1] dist = lca.dist(leaves[0], leaves[1]) for i in range(len(leaves)): for j in range(i + 1, len(leaves)): if dist != lca.dist(leaves[i], leaves[j]): print("No") exit() print("Yes") main()