結果
問題 | No.1637 Easy Tree Query |
ユーザー |
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提出日時 | 2025-03-20 20:56:54 |
言語 | PyPy3 (7.3.15) |
結果 |
AC
|
実行時間 | 349 ms / 2,000 ms |
コード長 | 1,509 bytes |
コンパイル時間 | 302 ms |
コンパイル使用メモリ | 82,656 KB |
実行使用メモリ | 101,456 KB |
最終ジャッジ日時 | 2025-03-20 20:57:09 |
合計ジャッジ時間 | 10,292 ms |
ジャッジサーバーID (参考情報) |
judge2 / judge3 |
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ファイルパターン | 結果 |
---|---|
sample | AC * 2 |
other | AC * 33 |
ソースコード
import sys from collections import deque def main(): n, q = map(int, sys.stdin.readline().split()) adj = [[] for _ in range(n+1)] for _ in range(n-1): a, b = map(int, sys.stdin.readline().split()) adj[a].append(b) adj[b].append(a) # BFS to build parent and children structure parent = [0] * (n+1) children = [[] for _ in range(n+1)] visited = [False] * (n+1) queue = deque([1]) visited[1] = True parent[1] = -1 # mark root's parent while queue: u = queue.popleft() for v in adj[u]: if not visited[v]: visited[v] = True parent[v] = u children[u].append(v) queue.append(v) # Calculate subtree sizes using iterative post-order traversal subtree_size = [1] * (n+1) stack = [(1, False)] while stack: node, visited_flag = stack.pop() if not visited_flag: stack.append((node, True)) # Add children in reverse to process them in order for child in reversed(children[node]): stack.append((child, False)) else: for child in children[node]: subtree_size[node] += subtree_size[child] # Process each query and print the sum total = 0 for _ in range(q): p, x = map(int, sys.stdin.readline().split()) total += subtree_size[p] * x print(total) if __name__ == "__main__": main()