結果
問題 |
No.3113 The farthest point
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ユーザー |
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提出日時 | 2025-04-19 03:25:32 |
言語 | PyPy3 (7.3.15) |
結果 |
WA
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実行時間 | - |
コード長 | 2,629 bytes |
コンパイル時間 | 535 ms |
コンパイル使用メモリ | 82,276 KB |
実行使用メモリ | 124,876 KB |
最終ジャッジ日時 | 2025-04-19 03:25:46 |
合計ジャッジ時間 | 14,267 ms |
ジャッジサーバーID (参考情報) |
judge1 / judge4 |
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ファイルパターン | 結果 |
---|---|
sample | AC * 3 |
other | AC * 1 WA * 32 |
ソースコード
# https://ikatakos.com/pot/programming_algorithm/graph_theory/lowest_common_ancestor class LcaDoubling: """ links[v] = { (u, w), (u, w), ... } (u:隣接頂点, w:辺の重み) というグラフ情報から、ダブリングによるLCAを構築。 任意の2頂点のLCAおよび距離を取得できるようにする """ def __init__(self, n, links, root=0): self.depths = [-1] * n self.distances = [-1] * n prev_ancestors = self._init_dfs(n, links, root) self.ancestors = [prev_ancestors] max_depth = max(self.depths) d = 1 while d < max_depth: next_ancestors = [prev_ancestors[p] for p in prev_ancestors] self.ancestors.append(next_ancestors) d <<= 1 prev_ancestors = next_ancestors def _init_dfs(self, n, links, root): q = [(root, -1, 0, 0)] direct_ancestors = [-1] * (n + 1) # 頂点数より1個長くし、存在しないことを-1で表す。末尾(-1)要素は常に-1 while q: v, p, dep, dist = q.pop() direct_ancestors[v] = p self.depths[v] = dep self.distances[v] = dist q.extend((u, v, dep + 1, dist + w) for u, w in links[v] if u != p) return direct_ancestors def get_lca(self, u, v): du, dv = self.depths[u], self.depths[v] if du > dv: u, v = v, u du, dv = dv, du tu = u tv = self.upstream(v, dv - du) if u == tv: return u for k in range(du.bit_length() - 1, -1, -1): mu = self.ancestors[k][tu] mv = self.ancestors[k][tv] if mu != mv: tu = mu tv = mv lca = self.ancestors[0][tu] assert lca == self.ancestors[0][tv] return lca def get_distance(self, u, v): lca = self.get_lca(u, v) return self.distances[u] + self.distances[v] - 2 * self.distances[lca] def upstream(self, v, k): i = 0 while k: if k & 1: v = self.ancestors[i][v] k >>= 1 i += 1 return v import sys input = sys.stdin.readline N = int(input()) G = [[] for _ in range(N)] for i in range(N-1): u, v, w = map(int, input().split()) u-=1 v-=1 G[u].append((v, w)) G[v].append((u, w)) LCA = LcaDoubling(N, G, 0) INF = 10**18 ma = -INF idx = -1 for i in range(1, N): d = LCA.get_distance(0, i) if ma<d: idx = i ans = 0 for i in range(N): d = LCA.get_distance(idx, i) ans = max(ans, d) print(ans)