結果

問題 No.2377 SUM AND XOR on Tree
コンテスト
ユーザー lam6er
提出日時 2025-03-31 17:55:43
言語 PyPy3
(7.3.15)
結果
AC  
実行時間 1,083 ms / 4,000 ms
コード長 2,447 bytes
コンパイル時間 287 ms
コンパイル使用メモリ 82,660 KB
実行使用メモリ 208,488 KB
最終ジャッジ日時 2025-03-31 17:57:08
合計ジャッジ時間 20,954 ms
ジャッジサーバーID
（参考情報） 		judge3 / judge5
このコードへのチャレンジ
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ファイルパターン 結果
sample AC * 2
other AC * 33
権限があれば一括ダウンロードができます

ソースコード

diff # 

import sys

MOD = 998244353

def main():
    sys.setrecursionlimit(1 << 25)
    N = int(sys.stdin.readline())
    edges = [[] for _ in range(N+1)]
    for _ in range(N-1):
        u, v = map(int, sys.stdin.readline().split())
        edges[u].append(v)
        edges[v].append(u)
    A = list(map(int, sys.stdin.readline().split()))
    A = [0] + A  # Convert to 1-based indexing

    # Build the tree structure with parent and children using BFS
    parent = [0] * (N + 1)
    children = [[] for _ in range(N + 1)]
    root = 1
    stack = [(root, 0)]
    while stack:
        u, p = stack.pop()
        parent[u] = p
        for v in edges[u]:
            if v != p:
                children[u].append(v)
                stack.append((v, u))

    ans = 0
    for b in range(30):
        # Precompute s_b for each node
        s = [0] * (N + 1)
        for u in range(1, N + 1):
            s[u] = (A[u] >> b) & 1
        
        dp0 = [0] * (N + 1)
        dp1 = [0] * (N + 1)
        
        # Iterative post-order traversal
        stack = [(root, False)]
        while stack:
            u, visited = stack.pop()
            if not visited:
                stack.append((u, True))
                # Push children in reverse order to process them left to right
                for v in reversed(children[u]):
                    stack.append((v, False))
            else:
                # Initialize current DP with the value of the node's s
                if s[u] == 0:
                    curr0, curr1 = 1, 0
                else:
                    curr0, curr1 = 0, 1
                # Merge each child's DP
                for v in children[u]:
                    v0 = dp0[v]
                    v1 = dp1[v]
                    # Cut the edge to child (contributes v's 1)
                    new0_cut = curr0 * v1 % MOD
                    new1_cut = curr1 * v1 % MOD
                    # Not cut the edge (merge with child's parity)
                    nc0 = (curr0 * v0 + curr1 * v1) % MOD
                    nc1 = (curr0 * v1 + curr1 * v0) % MOD
                    # Update current DP
                    curr0 = (new0_cut + nc0) % MOD
                    curr1 = (new1_cut + nc1) % MOD
                dp0[u] = curr0
                dp1[u] = curr1
        
        # Add the contribution of this bit
        ans = (ans + (dp1[root] % MOD) * ((1 << b) % MOD)) % MOD

    print(ans % MOD)

if __name__ == '__main__':
    main()
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