結果

問題 No.1928 Make a Binary Tree
ユーザー ecottea
提出日時 2022-05-06 21:45:48
言語 C++14
(gcc 13.3.0 + boost 1.87.0)
結果
AC  
実行時間 133 ms / 3,000 ms
コード長 7,969 bytes
コンパイル時間 4,047 ms
コンパイル使用メモリ 235,200 KB
実行使用メモリ 48,496 KB
最終ジャッジ日時 2024-07-05 22:56:33
合計ジャッジ時間 9,585 ms
ジャッジサーバーID
(参考情報)
judge4 / judge3
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ファイルパターン 結果
sample AC * 3
other AC * 57
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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

#ifndef HIDDEN_IN_VS //
//
#define _CRT_SECURE_NO_WARNINGS
//
#include <bits/stdc++.h>
using namespace std;
//
using ll = long long; // -2^63 2^63 = 9 * 10^18int -2^31 2^31 = 2 * 10^9
using pii = pair<int, int>; using pll = pair<ll, ll>; using pil = pair<int, ll>; using pli = pair<ll, int>;
using vi = vector<int>; using vvi = vector<vi>; using vvvi = vector<vvi>;
using vl = vector<ll>; using vvl = vector<vl>; using vvvl = vector<vvl>;
using vb = vector<bool>; using vvb = vector<vb>; using vvvb = vector<vvb>;
using vc = vector<char>; using vvc = vector<vc>; using vvvc = vector<vvc>;
using vd = vector<double>; using vvd = vector<vd>; using vvvd = vector<vvd>;
template <class T> using priority_queue_rev = priority_queue<T, vector<T>, greater<T>>;
using Graph = vvi;
//
const double PI = acos(-1);
const vi dx4 = { 1, 0, -1, 0 }; // 4
const vi dy4 = { 0, 1, 0, -1 };
const vi dx8 = { 1, 1, 0, -1, -1, -1, 0, 1 }; // 8
const vi dy8 = { 0, 1, 1, 1, 0, -1, -1, -1 };
const int INF = 1001001001; const ll INFL = 4004004004004004004LL;
const double EPS = 1e-12; // 調
//
struct fast_io { fast_io() { cin.tie(nullptr); ios::sync_with_stdio(false); cout << fixed << setprecision(15); } } fastIOtmp;
//
#define all(a) (a).begin(), (a).end()
#define sz(x) ((int)(x).size())
#define distance (int)distance
#define Yes(b) {cout << ((b) ? "Yes\n" : "No\n");}
#define rep(i, n) for(int i = 0, i##_len = int(n); i < i##_len; ++i) // 0 n-1
#define repi(i, s, t) for(int i = int(s), i##_end = int(t); i <= i##_end; ++i) // s t
#define repir(i, s, t) for(int i = int(s), i##_end = int(t); i >= i##_end; --i) // s t
#define repe(v, a) for(const auto& v : (a)) // a
#define repea(v, a) for(auto& v : (a)) // a
#define repb(set, d) for(int set = 0; set < (1 << int(d)); ++set) // d
#define repp(a) sort(all(a)); for(bool a##_perm = true; a##_perm; a##_perm = next_permutation(all(a))) // a
#define smod(n, m) ((((n) % (m)) + (m)) % (m)) // mod
#define uniq(a) {sort(all(a)); (a).erase(unique(all(a)), (a).end());} //
#define EXIT(a) {cout << (a) << endl; exit(0);} //
//
template <class T> inline ll pow(T n, int k) { ll v = 1; rep(i, k) v *= n; return v; }
template <class T> inline bool chmax(T& M, const T& x) { if (M < x) { M = x; return true; } return false; } // true
    
template <class T> inline bool chmin(T& m, const T& x) { if (m > x) { m = x; return true; } return false; } // true
    
//
template <class T, class U> inline istream& operator>>(istream& is, pair<T, U>& p) { is >> p.first >> p.second; return is; }
template <class T> inline istream& operator>>(istream& is, vector<T>& v) { repea(x, v) is >> x; return is; }
template <class T> inline vector<T>& operator--(vector<T>& v) { repea(x, v) --x; return v; }
template <class T> inline vector<T>& operator++(vector<T>& v) { repea(x, v) ++x; return v; }
// Visual Studio
#ifdef _MSC_VER
#include "local.hpp"
// gcc
#else
#define popcount (int)__builtin_popcount
#define popcountll (int)__builtin_popcountll
#define lsb __builtin_ctz
#define lsbll __builtin_ctzll
#define msb(n) (31 - __builtin_clz(n))
#define msbll(n) (63 - __builtin_clzll(n))
#define gcd __gcd
#define dump(...)
#define dumpel(v)
#define input_from_file(f)
#define output_to_file(f)
#endif
#endif //
//--------------AtCoder --------------
#include <atcoder/all>
using namespace atcoder;
//using mint = modint1000000007;
using mint = modint998244353;
//using mint = modint; // mint::set_mod(m);
istream& operator>>(istream& is, mint& x) { ll x_; is >> x_; x = x_; return is; }
ostream& operator<<(ostream& os, const mint& x) { os << x.val(); return os; }
using vm = vector<mint>; using vvm = vector<vm>; using vvvm = vector<vvm>;
//----------------------------------------
//O(|V| + |E|)
/*
* n m
*
* n :
* m :
* g :
* undirected : true
* one_indexed : 1-indexed true
*/
void read_graph(int n, int m, Graph& g, bool undirected = true, bool one_indexed = true) {
g = Graph(n);
rep(i, m) {
int a, b;
cin >> a >> b;
if (one_indexed) { a--; b--; }
g[a].push_back(b);
if (undirected) g[b].push_back(a);
}
}
//
/*
* Skew_heap<T>() : O(1)
* T
*
* bool empty() : O(1)
*
*
* int size() : O(1)
*
*
* T top() : O(1)
*
*
* push(T val) : O(log n)
* val
*
* pop() : O(log n)
*
*
* merge(Skew_heap<T>& hp) : O(log n)
* hp
*/
template <typename T> struct Skew_heap {
// : https://kopricky.github.io/code/DataStructure_Advanced/skew_heap.html
// verify : https://atcoder.jp/contests/abc246/tasks/abc246_g
struct Node {
Node* l, * r;
T v;
Node(T v_) : l(nullptr), r(nullptr), v(v_) {}
friend Node* meld(Node* a, Node* b) {
if (a == nullptr) return b;
if (b == nullptr) return a;
// a >= b
if (a->v < b->v) swap(a, b);
// b a
a->l = meld(a->l, b);
//
swap(a->l, a->r);
return a;
}
};
Node* root;
int n;
//
Skew_heap() : root(nullptr), n(0) {}
//
bool empty() const { return root == nullptr; }
//
int size() const { return n; }
// val
void push(T val) {
Node* p = new Node(val);
root = meld(root, p);
n++;
}
//
T top() const { return root->v; }
//
void pop() {
Node* p = root;
root = meld(root->r, root->l);
delete p;
n--;
}
// hp
void merge(Skew_heap<T>& hp) {
n += hp.n;
root = meld(root, hp.root);
hp.root = nullptr;
}
#ifdef _MSC_VER
friend ostream& operator<<(ostream& os, const Skew_heap<T>& q) {
q.print_rf(os, q.root);
return os;
}
void print_rf(ostream& os, const Node* pt) const {
if (pt == nullptr) return;
os << pt->v << " "; print_rf(os, pt->l); print_rf(os, pt->r);
}
#endif
};
int main() {
// input_from_file("input.txt");
// output_to_file("output.txt");
int n;
cin >> n;
Graph g;
read_graph(n, n - 1, g);
vector<Skew_heap<int>> qs(n);
//
function<void(int, int)> dfs = [&](int s, int p) {
repe(t, g[s]) {
//
if (t == p) continue;
//
dfs(t, s);
qs[s].merge(qs[t]);
}
int w = 1;
if (!qs[s].empty()) {
w += qs[s].top();
qs[s].pop();
if (!qs[s].empty()) {
w += qs[s].top();
qs[s].pop();
}
}
qs[s].push(w);
};
//
dfs(0, -1);
cout << qs[0].top() << endl;
}
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