結果
| 問題 | 
                            No.1094 木登り / Climbing tree
                             | 
                    
| ユーザー | 
                             | 
                    
| 提出日時 | 2022-11-04 00:14:12 | 
| 言語 | PyPy3  (7.3.15)  | 
                    
| 結果 | 
                             
                                TLE
                                 
                             
                            
                         | 
                    
| 実行時間 | - | 
| コード長 | 1,225 bytes | 
| コンパイル時間 | 156 ms | 
| コンパイル使用メモリ | 82,432 KB | 
| 実行使用メモリ | 161,664 KB | 
| 最終ジャッジ日時 | 2024-07-18 05:55:09 | 
| 合計ジャッジ時間 | 47,900 ms | 
| 
                            ジャッジサーバーID (参考情報)  | 
                        judge2 / judge4 | 
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| ファイルパターン | 結果 | 
|---|---|
| sample | AC * 1 | 
| other | AC * 11 TLE * 15 | 
ソースコード
from collections import deque
n = int(input())
g = [[] for _ in range(n)]
for _ in range(n - 1):
  a, b, c = map(int, input().split())
  a -= 1
  b -= 1
  g[a].append((b, c))
  g[b].append((a, c))
k = n.bit_length()
src = 0
doub = [[src for _ in range(k)] for _ in range(n)]
depth = [None for _ in range(n)]
depth[src] = 0
cost = [0 for _ in range(n)]
dq = deque()
dq.appendleft(src)
while len(dq) > 0:
  cur = dq.pop()
  for nxt, c in g[cur]:
    if not depth[nxt] is None:
      continue
    depth[nxt] = depth[cur] + 1
    cost[nxt] = cost[cur] + c
    doub[nxt][0] = cur
    dq.appendleft(nxt)
for i in range(1, k):
  for v in range(n):
    doub[v][i] = doub[doub[v][i - 1]][i - 1]
def lca(u, v):
  ut = u
  vt = v
  if depth[ut] > depth[vt]:
    ut, vt = vt, ut
  diff = depth[vt] - depth[ut]
  for i in range(k):
    if (diff >> i) & 1 != 0:
      vt = doub[vt][i]
  if ut != vt:
    for i in range(k - 1, -1, -1):
      if doub[ut][i] != doub[vt][i]:
        ut = doub[ut][i]
        vt = doub[vt][i]
    
    if ut != vt:
      ut = doub[ut][0]
  
  return ut
q = int(input())
for _ in range(q):
  s, t = map(int, input().split())
  s -= 1
  t -= 1
  l = lca(s, t)
  print(cost[s] + cost[t] - 2 * cost[l])