結果

問題 No.2504 NOT Path Painting
ユーザー suisensuisen
提出日時 2023-02-21 08:47:21
言語 C++17
(gcc 13.3.0 + boost 1.87.0)
結果
RE  
(最新)
AC  
(最初)
実行時間 -
コード長 4,123 bytes
コンパイル時間 1,117 ms
コンパイル使用メモリ 98,404 KB
最終ジャッジ日時 2025-02-10 19:38:21
ジャッジサーバーID
(参考情報)
judge3 / judge4
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
other RE * 8 MLE * 13
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

#include <deque>
#include <iostream>
#include <tuple>
#include <vector>
#include <atcoder/modint>
using mint = atcoder::modint998244353;
int edge_num(int n) {
return (n * (n + 1)) >> 1;
}
int main() {
std::ios::sync_with_stdio(false);
std::cin.tie(nullptr);
int n;
std::cin >> n;
std::vector<std::vector<int>> g(n);
for (int i = 0; i < n - 1; ++i) {
int u, v;
std::cin >> u >> v;
--u, --v;
g[u].push_back(v);
g[v].push_back(u);
}
const int m = edge_num(n);
const mint inv_m = mint(m).inv();
struct SubtreeSize {
SubtreeSize(int n, const std::vector<std::vector<int>> &g) : _n(n), _par(_n, -1), _siz(_n, 1) {
auto dfs = [&](auto dfs, int u, int p) -> int {
_par[u] = p;
for (int v : g[u]) if (v != p) {
_siz[u] += dfs(dfs, v, u);
}
return _siz[u];
};
dfs(dfs, 0, -1);
}
int operator()(int u, int p) const {
return _par[u] == p ? _siz[u] : _n - _siz[p];
}
int t(int u, int ng1) const {
return _n - (*this)(ng1, u);
}
int t(int u, int ng1, int ng2) const {
return _n - (*this)(ng1, u) - (*this)(ng2, u);
}
private:
int _n;
std::vector<int> _par, _siz;
} subtree_size { n, g };
std::vector<mint> ans_f(n, 0);
for (int x = 0; x < n; ++x) {
int u_x = edge_num(n);
for (int y : g[x]) {
u_x -= edge_num(subtree_size(y, x));
}
ans_f[x] = m * mint(m - u_x).inv();
}
std::vector<std::vector<mint>> ans_g(n, std::vector<mint>(n));
std::vector<std::vector<int>> par(n, std::vector<int>(n, -1));
// x, y, A, B
std::deque<std::tuple<int, int, mint, mint>> dq;
for (int x = 0; x < n; ++x) {
ans_g[x][x] = ans_f[x];
int u_x = edge_num(n);
for (int y : g[x]) {
const int s_y = subtree_size(y, x);
u_x -= edge_num(s_y);
}
for (int y : g[x]) {
const int s_y = subtree_size(y, x);
par[x][y] = x;
const int u_y = u_x - s_y * (n - s_y);
const mint A = u_y * ans_f[x];
const mint B = 0;
dq.emplace_back(x, y, A, B);
}
}
while (dq.size()) {
auto [x, z, A, B] = dq.front();
dq.pop_front();
const int par_z = par[x][z];
const int s_z = subtree_size(z, par_z);
int u_z = edge_num(s_z);
for (int y : g[z]) if (y != par_z) {
u_z -= edge_num(subtree_size(y, z));
}
const int t_z = s_z;
ans_g[x][z] = A + u_z * ans_f[z] + B;
int prev_z2 = z, z2 = par_z;
while (z2 != x) {
const int next_z2 = par[x][z2];
const int t_z2 = subtree_size.t(z2, prev_z2, next_z2);
ans_g[x][z] += t_z * t_z2 * ans_g[z][z2];
std::tie(prev_z2, z2) = std::make_tuple(z2, next_z2);
}
const int t_x = subtree_size.t(x, prev_z2);
ans_g[x][z] = (1 + ans_g[x][z] * inv_m) * (1 - t_x * t_z * inv_m).inv();
for (int y : g[z]) if (y != par_z) {
const int s_y = subtree_size(y, z);
const int next_t_z = t_z - s_y;
const int u_y = u_z - s_y * (s_z - s_y);
const mint next_A = A + u_y * ans_f[z];
mint next_B = B + next_t_z * t_x * ans_g[x][z];
int prev_z2 = z, z2 = par_z;
while (z2 != x) {
const int next_z2 = par[x][z2];
const int t_z2 = subtree_size.t(z2, prev_z2, next_z2);
next_B += next_t_z * t_z2 * ans_g[z][z2];
std::tie(prev_z2, z2) = std::make_tuple(z2, next_z2);
}
par[x][y] = z;
dq.emplace_back(x, y, next_A, next_B);
}
}
mint ans = 1;
for (int x = 0; x < n; ++x) {
for (int y = 0; y <= x; ++y) {
ans += ans_g[x][y] * inv_m;
}
}
std::cout << ans.val() << '\n';
return 0;
}
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