結果
問題 | No.1507 Road Blocked |
ユーザー | srjywrdnprkt |
提出日時 | 2023-06-18 22:48:59 |
言語 | C++17 (gcc 12.3.0 + boost 1.83.0) |
結果 |
AC
|
実行時間 | 126 ms / 2,000 ms |
コード長 | 5,495 bytes |
コンパイル時間 | 2,366 ms |
コンパイル使用メモリ | 213,336 KB |
実行使用メモリ | 20,096 KB |
最終ジャッジ日時 | 2024-06-26 09:26:17 |
合計ジャッジ時間 | 8,311 ms |
ジャッジサーバーID (参考情報) |
judge4 / judge2 |
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テストケース
テストケース表示入力 | 結果 | 実行時間 実行使用メモリ |
---|---|---|
testcase_00 | AC | 2 ms
5,248 KB |
testcase_01 | AC | 2 ms
5,376 KB |
testcase_02 | AC | 2 ms
5,376 KB |
testcase_03 | AC | 84 ms
20,096 KB |
testcase_04 | AC | 123 ms
17,152 KB |
testcase_05 | AC | 120 ms
17,132 KB |
testcase_06 | AC | 120 ms
17,280 KB |
testcase_07 | AC | 121 ms
17,280 KB |
testcase_08 | AC | 117 ms
17,280 KB |
testcase_09 | AC | 120 ms
17,152 KB |
testcase_10 | AC | 119 ms
17,280 KB |
testcase_11 | AC | 120 ms
17,152 KB |
testcase_12 | AC | 119 ms
17,152 KB |
testcase_13 | AC | 120 ms
17,212 KB |
testcase_14 | AC | 123 ms
17,280 KB |
testcase_15 | AC | 120 ms
17,152 KB |
testcase_16 | AC | 121 ms
17,152 KB |
testcase_17 | AC | 120 ms
17,152 KB |
testcase_18 | AC | 120 ms
17,280 KB |
testcase_19 | AC | 121 ms
17,152 KB |
testcase_20 | AC | 120 ms
17,152 KB |
testcase_21 | AC | 121 ms
17,152 KB |
testcase_22 | AC | 122 ms
17,280 KB |
testcase_23 | AC | 126 ms
17,280 KB |
testcase_24 | AC | 124 ms
17,152 KB |
testcase_25 | AC | 120 ms
17,280 KB |
testcase_26 | AC | 124 ms
17,280 KB |
testcase_27 | AC | 122 ms
17,280 KB |
testcase_28 | AC | 124 ms
17,152 KB |
testcase_29 | AC | 126 ms
17,280 KB |
testcase_30 | AC | 126 ms
17,152 KB |
testcase_31 | AC | 125 ms
17,152 KB |
testcase_32 | AC | 125 ms
17,152 KB |
ソースコード
#include <bits/stdc++.h> using namespace std; using ll = long long; template<typename S> struct Tree{ vector<vector<S>> E, par0; vector<S> dist0; S N, log=0; Tree (const vector<vector<S>> &_E){ N = _E.size(); E = _E; } //fromを根とする木の各頂点の深さを求める vector<S> depth (S from) { vector<S> dist(N); _depth(from, -1, dist); return dist; } void _depth(S from, S p, vector<S> &dist) { for (auto to : E[from]){ if (to == p) continue; dist[to] = dist[from]+1; _depth(to, from, dist); } } //木の二頂点(a, b)間の最短距離を求める S dist(S a, S b){ S c = lca(a, b); return dist0[a] + dist0[b]- 2*dist0[c]; } //木の二頂点(a, b)のLCAを求める S lca(S a, S b){ if (par0.size() == 0){ dist0 = depth(0); _doubling(); } if (dist0[a] < dist0[b]) swap(a, b); for (S i=0; i<=log; i++){ if ((dist0[a]-dist0[b]) & (1LL<<i)) a = par0[i][a]; } if (a == b) return a; for (S i=log; i>=0; i--){ if (par0[i][a] != par0[i][b]){ a = par0[i][a]; b = par0[i][b]; } } return par0[0][a]; } void _doubling(){ S cnt = 1; while(cnt < N){ cnt *= 2; log++; } par0.resize(log+1, vector<S>(N)); _ancestor(0, -1); for (S i=1; i<=log; i++){ for (S j=0; j<N; j++){ if (par0[i-1][j] == -1) par0[i][j] = -1; else par0[i][j] = par0[i-1][par0[i-1][j]]; } } } void _ancestor(S from, S p){ par0[0][from] = p; for (auto to : E[from]){ if (to == p) continue; _ancestor(to, from); } } //fromとgoalの最短経路上に含まれる点を全て求める vector<S> shortest_path(S from, S goal) const{ vector<S> path, pt; _shortest_path(from, goal, -1, pt, path); return path; } void _shortest_path(S from, S goal, S p, vector<S> &pt, vector<S> &path) const{ pt.push_back(from); if (from == goal) path = pt; for (auto to : E[from]){ if (to == p) continue; _shortest_path(to, goal, from, pt, path); } pt.pop_back(); } //木の直径とその両端の点を求める tuple<S, S, S> diameter() const{ S s=0, t=0, mx=0; _diameter(s, -1, 0, mx, t); s=t; t=0; mx=0; _diameter(s, -1, 0, mx, t); return make_tuple(s, t, mx); } void _diameter(S from, S p, S d, S &mx, S &argmx) const{ if (d > mx){ argmx = from; mx = d; } for (auto to : E[from]){ if (to == p) continue; _diameter(to, from, d+1, mx, argmx); } } //fromを根とする木の部分木のサイズを求める vector<S> subtree_size(S from) const{ vector<S> subtree(N); _subtree_size(from, -1, subtree); return subtree; } S _subtree_size(S from, S p, vector<S> &subtree) const{ S cnt = 1; for (auto to : E[from]){ if (to == p) continue; cnt += _subtree_size(to, from, subtree); } return subtree[from] = cnt; } }; const ll modc = 998244353; class mint { ll x; public: mint(ll x=0) : x((x%modc+modc)%modc) {} mint operator-() const { return mint(-x); } mint& operator+=(const mint& a) { if ((x += a.x) >= modc) x -= modc; return *this; } mint& operator-=(const mint& a) { if ((x += modc-a.x) >= modc) x -= modc; return *this; } mint& operator*=(const mint& a) { (x *= a.x) %= modc; return *this; } mint operator+(const mint& a) const { mint res(*this); return res+=a; } mint operator-(const mint& a) const { mint res(*this); return res-=a; } mint operator*(const mint& a) const { mint res(*this); return res*=a; } mint pow(ll t) const { if (!t) return 1; mint a = pow(t>>1); a *= a; if (t&1) a *= *this; return a; } mint inv() const { return pow(modc-2); } mint& operator/=(const mint& a) { return (*this) *= a.inv(); } mint operator/(const mint& a) const { mint res(*this); return res/=a; } bool operator == (const mint& a) const{ return x == a.x; } friend ostream& operator<<(ostream& os, const mint& m){ os << m.x; return os; } friend istream& operator>>(istream& ip, mint &m) { ll t; ip >> t; m = mint(t); return ip; } }; int main(){ ll N, C; mint ans=0; cin >> N; vector<ll> A(N-1), B(N-1); vector<vector<long long>> E(N); for (int i=0; i < N-1; i++){ cin >> A[i] >> B[i]; A[i]--; B[i]--; E[A[i]].push_back(B[i]); E[B[i]].push_back(A[i]); } Tree tree(E); vector<ll> subtree = tree.subtree_size(0); for (int i=0; i<N-1; i++){ C = min(subtree[A[i]], subtree[B[i]]); ans += C * (N-C); } ans /= N; ans *= 2; ans /= (N-1); ans /= (N-1); cout << -ans+1 << endl; return 0; }