結果
| 問題 |
No.1637 Easy Tree Query
|
| コンテスト | |
| ユーザー |
lam6er
|
| 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()
lam6er