結果

問題 No.2504 NOT Path Painting
ユーザー suisen
提出日時 2023-02-21 09:46:43
言語 C++17
(gcc 13.3.0 + boost 1.87.0)
結果
RE  
実行時間 -
コード長 3,928 bytes
コンパイル時間 3,992 ms
コンパイル使用メモリ 93,800 KB
最終ジャッジ日時 2025-02-10 19:38:46
ジャッジサーバーID
(参考情報)
judge1 / 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
other RE * 8 MLE * 13
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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
std::deque<std::tuple<int, int>> dq;
for (int x = 0; x < n; ++x) {
ans_g[x][x] = ans_f[x];
for (int y : g[x]) {
par[x][y] = x;
dq.emplace_back(x, y);
}
}
while (dq.size()) {
auto [x, z] = dq.front();
dq.pop_front();
std::vector<int> Pxz;
for (int a = z; a != -1; a = par[x][a]) {
Pxz.push_back(a);
}
assert(Pxz.front() == z and Pxz.back() == x and Pxz.size() >= 2);
const int l = Pxz.size();
auto get_t_z = [&](int idx) {
return
idx == 0 ? subtree_size.t(Pxz[idx], Pxz[idx + 1])
: idx == l - 1 ? subtree_size.t(Pxz[idx], Pxz[idx - 1])
: subtree_size.t(Pxz[idx], Pxz[idx - 1], Pxz[idx + 1]);
};
for (int i = 0; i < l; ++i) {
const int z2 = Pxz[i];
std::vector<int> ng;
if (i > 0) ng.push_back(Pxz[i - 1]);
if (i + 1 < l) ng.push_back(Pxz[i + 1]);
int u_z2 = 1;
{
int sum = 0;
for (int nz : g[z2]) if (nz != ng.front() and nz != ng.back()) {
int s_nz = subtree_size(nz, z2);
u_z2 += sum * s_nz;
sum += s_nz;
}
u_z2 += sum;
}
ans_g[x][z] += u_z2 * ans_f[z2];
const int t2 = get_t_z(i);
for (int j = 0; j < i; ++j) {
if (i == l - 1 and j == 0) continue;
const int z1 = Pxz[j];
const int t1 = get_t_z(j);
ans_g[x][z] += t1 * t2 * ans_g[z2][z1];
}
}
ans_g[x][z] = (1 + ans_g[x][z] * inv_m) * (1 - get_t_z(l - 1) * get_t_z(0) * inv_m).inv();
for (int y : g[z]) if (y != par[x][z]) {
par[x][y] = z;
dq.emplace_back(x, y);
}
}
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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