結果
問題 | No.2504 NOT Path Painting |
ユーザー |
|
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 |
(要ログイン)
ファイルパターン | 結果 |
---|---|
other | RE * 8 MLE * 13 |
ソースコード
#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, Bstd::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;}