結果
問題 |
No.1038 TreeAddQuery
|
ユーザー |
![]() |
提出日時 | 2025-04-15 21:07:32 |
言語 | PyPy3 (7.3.15) |
結果 |
TLE
|
実行時間 | - |
コード長 | 2,400 bytes |
コンパイル時間 | 377 ms |
コンパイル使用メモリ | 82,096 KB |
実行使用メモリ | 78,880 KB |
最終ジャッジ日時 | 2025-04-15 21:13:29 |
合計ジャッジ時間 | 8,894 ms |
ジャッジサーバーID (参考情報) |
judge4 / judge5 |
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ファイルパターン | 結果 |
---|---|
sample | AC * 3 |
other | AC * 5 TLE * 1 -- * 18 |
ソースコード
import sys from collections import deque sys.setrecursionlimit(1 << 25) def main(): input = sys.stdin.read().split() ptr = 0 N, Q = int(input[ptr]), int(input[ptr+1]) ptr += 2 adj = [[] for _ in range(N+1)] for _ in range(N-1): a = int(input[ptr]) b = int(input[ptr+1]) adj[a].append(b) adj[b].append(a) ptr += 2 LOG = 20 parent = [[-1]*(N+1) for _ in range(LOG)] depth = [0]*(N+1) root = 1 q = deque([root]) parent[0][root] = -1 while q: u = q.popleft() for v in adj[u]: if parent[0][v] == -1 and v != parent[0][u]: parent[0][v] = u depth[v] = depth[u] + 1 q.append(v) for k in range(1, LOG): for v in range(1, N+1): if parent[k-1][v] != -1: parent[k][v] = parent[k-1][parent[k-1][v]] def lca(u, v): if depth[u] < depth[v]: u, v = v, u for k in range(LOG-1, -1, -1): if depth[u] - (1 << k) >= depth[v]: u = parent[k][u] if u == v: return u for k in range(LOG-1, -1, -1): if parent[k][u] != parent[k][v]: u = parent[k][u] v = parent[k][v] return parent[0][u] def distance(u, v): a = lca(u, v) return depth[u] + depth[v] - 2 * depth[a] K = 300 ans_small = [0]*(N+1) large_queries = [] visited = [0]*(N+1) timestamp = 0 for _ in range(Q): X_i = int(input[ptr]) Y_i = int(input[ptr+1]) Z_i = int(input[ptr+2]) ptr += 3 current_sum = ans_small[X_i] for (X_j, Y_j, Z_j) in large_queries: if Y_j >= distance(X_j, X_i): current_sum += Z_j print(current_sum) if Y_i <= K: timestamp += 1 q = deque() q.append((X_i, 0)) visited[X_i] = timestamp while q: u, dist = q.popleft() ans_small[u] += Z_i if dist < Y_i: for v in adj[u]: if visited[v] != timestamp: visited[v] = timestamp q.append((v, dist + 1)) else: large_queries.append((X_i, Y_i, Z_i)) if __name__ == "__main__": main()