結果

問題 No.1103 Directed Length Sum
ユーザー terasa
提出日時 2022-11-20 12:55:19
言語 PyPy3
(7.3.15)
結果
TLE  
実行時間 -
コード長 3,620 bytes
コンパイル時間 249 ms
コンパイル使用メモリ 81,692 KB
実行使用メモリ 410,468 KB
最終ジャッジ日時 2024-09-21 13:27:11
合計ジャッジ時間 14,684 ms
ジャッジサーバーID
（参考情報） 		judge3 / judge2
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 * 21 TLE * 1
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 #
プレゼンテーションモードにする

from typing import List, Tuple, Callable, TypeVar
from typing import List, Tuple, Optional
import sys
import itertools
import heapq
import bisect
import math
from collections import deque, defaultdict
from functools import lru_cache, cmp_to_key
input = sys.stdin.readline
if __file__ != 'prog.py':
    sys.setrecursionlimit(10 ** 6)
def readints(): return map(int, input().split())
def readlist(): return list(readints())
def readstr(): return input()[:-1]
T = TypeVar('T')
input = sys.stdin.readline
class Rerooting:
    # reference: https://null-mn.hatenablog.com/entry/2020/04/14/124151
    # 適当な頂点vを根とする部分木に対して計算される値dp_vが、vの子c1, c2, ... ckを用いて
    # 下記のように表すことができる
    # dp_v = g(merge(f(dp_c1,c1), f(dp_c2,c2), ..., f(dp_ck,ck)), v)
    def __init__(self, N: int, E: List[Tuple[int, int]],
                 f: Callable[[T, int, int, int], T],
                 g: Callable[[T, int], T],
                 merge: Callable[[T, T], T], e: T,
                 root: int = 0):
        self.N = N
        self.E = E
        self.f = f
        self.g = g
        self.merge = merge
        self.e = e
        self.dp = [[self.e for _ in range(len(self.E[v]))] for v in range(self.N)]
        self._calculate(root)
    def _dfs1(self, root):
        stack = [(root, -1)]
        ret = [self.e] * self.N
        while stack:
            v, p = stack.pop()
            if v < 0:
                v = ~v
                acc = self.e
                for i, (c, d) in enumerate(self.E[v]):
                    if d == p:
                        continue
                    self.dp[v][i] = ret[d]
                    acc = self.merge(acc, self.f(ret[d], v, d, c))
                ret[v] = self.g(acc, v)
                continue
            stack.append((~v, p))
            for i, (c, d) in enumerate(self.E[v]):
                if d == p:
                    continue
                stack.append((d, v))
    def _dfs2(self, root):
        stack = [(root, -1, self.e)]
        while stack:
            v, p, from_par = stack.pop()
            for i, (c, d) in enumerate(self.E[v]):
                if d == p:
                    self.dp[v][i] = from_par
                    break
            ch = len(self.E[v])
            Sr = [self.e] * (ch + 1)
            for i in range(ch, 0, -1):
                c, d = self.E[v][i - 1]
                Sr[i - 1] = self.merge(Sr[i], self.f(self.dp[v][i - 1], v, d, c))
            Sl = self.e
            for i, (c, d) in enumerate(self.E[v]):
                if d != p:
                    val = self.merge(Sl, Sr[i + 1])
                    stack.append((d, v, self.g(val, v)))
                Sl = self.merge(Sl, self.f(self.dp[v][i], v, d, c))
    def _calculate(self, root):
        self._dfs1(root)
        self._dfs2(root)
    def solve(self, v):
        ans = self.e
        for i, (c, d) in enumerate(self.E[v]):
            ans = self.merge(ans, self.f(self.dp[v][i], v, d, c))
        return self.g(ans, v)
N = int(input())
E = [[] for _ in range(N)]
D = [0] * N
for _ in range(N - 1):
    a, b = readints()
    a -= 1
    b -= 1
    E[a].append((1, b))
    D[b] += 1
for i in range(N):
    if D[i] == 0:
        root = i
        break
mod = 10 ** 9 + 7
def f(a, v, ch, cost):
    return ((a[0] + a[1]) % mod, a[1])
def g(a, v):
    return (a[0], a[1] + 1)
def merge(a, b):
    return ((a[0] + b[0]) % mod, a[1] + b[1])
solver = Rerooting(N, E, f, g, merge, (0, 0), root=root)
ans = 0
for i in range(N):
    ans += solver.solve(i)[0]
    ans %= mod
print(ans)
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