結果
問題 | No.778 クリスマスツリー |
ユーザー |
![]() |
提出日時 | 2021-10-20 12:45:30 |
言語 | C++14 (gcc 13.3.0 + boost 1.87.0) |
結果 |
AC
|
実行時間 | 864 ms / 2,000 ms |
コード長 | 13,825 bytes |
コンパイル時間 | 2,066 ms |
コンパイル使用メモリ | 145,296 KB |
実行使用メモリ | 103,044 KB |
最終ジャッジ日時 | 2024-09-20 06:32:50 |
合計ジャッジ時間 | 8,674 ms |
ジャッジサーバーID (参考情報) |
judge4 / judge1 |
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ファイルパターン | 結果 |
---|---|
sample | AC * 3 |
other | AC * 12 |
ソースコード
/* #region header */#pragma GCC optimize("O3") //コンパイラ最適化用#ifdef LOCAL#define _GLIBCXX_DEBUG //配列に[]でアクセス時のエラー表示#endif#define _USE_MATH_DEFINES#include <algorithm> //sort,二分探索,など#include <bitset> //固定長bit集合// #include <boost/multiprecision/cpp_dec_float.hpp>// #include <boost/multiprecision/cpp_int.hpp>#include <cassert> //assert(p)で,not pのときにエラー#include <cctype>#include <chrono> //実行時間計測#include <climits>#include <cmath> //pow,logなど#include <complex> //複素数#include <cstdio>#include <cstring>#include <deque>#include <functional> //sortのgreater#include <iomanip> //setprecision(浮動小数点の出力の誤差)#include <ios> // std::left, std::right#include <iostream> //入出力#include <iterator> //集合演算(積集合,和集合,差集合など)#include <map>#include <numeric> //iota(整数列の生成),gcdとlcm,accumulate#include <queue>#include <random>#include <set>#include <stack>#include <string>#include <unordered_map>#include <unordered_set>#include <utility> //pair#include <vector>using namespace std;typedef long long LL;typedef long double LD;#define ALL(x) x.begin(), x.end()const long long INF = numeric_limits<long long>::max() / 4;const int MOD = 1e9 + 7;// const int MOD=998244353;//略記#define FF first#define SS second#define int long long#define stoi stoll#define LD long double#define PII pair<int, int>#define PB push_back#define EB emplace_back#define MP make_pair#define SZ(x) (int)((x).size())#define VB vector<bool>#define VVB vector<vector<bool>>#define VI vector<int>#define VVI vector<vector<int>>#define REP(i, n) for (int i = 0; i < (int)(n); i++)#define REPD(i, n) for (int i = (int)(n) - (int)1; i >= 0; i--)#define FOR(i, a, b) for (int i = a; i < (int)(b); i++)#define FORD(i, a, b) for (int i = (int)(b) - (int)1; i >= (int)a; i--)const int dx[4] = {0, 1, 0, -1}, dy[4] = {-1, 0, 1, 0};const int Dx[8] = {0, 1, 1, 1, 0, -1, -1, -1},Dy[8] = {-1, -1, 0, 1, 1, 1, 0, -1};int in() {int x;cin >> x;return x;}// https://qiita.com/Lily0727K/items/06cb1d6da8a436369eed#c%E3%81%A7%E3%81%AE%E5%AE%9F%E8%A3%85void print() { cout << "\n"; }template <class Head, class... Tail> void print(Head &&head, Tail &&...tail) {cout << head;if (sizeof...(tail) != 0)cout << " ";print(forward<Tail>(tail)...);}template <class T> void print(vector<T> &vec) {for (auto &a : vec) {cout << a;if (&a != &vec.back())cout << " ";}cout << "\n";}template <class T> void print(set<T> &set) {for (auto &a : set) {cout << a << " ";}cout << "\n";}template <class T> void print(vector<vector<T>> &df) {for (auto &vec : df) {print(vec);}}// debug macro// https://atcoder.jp/contests/abc202/submissions/22815715namespace debugger {template <class T> void view(const std::vector<T> &a) {std::cerr << "{ ";for (const auto &v : a) {std::cerr << v << ", ";}std::cerr << "\b\b }";}template <class T> void view(const std::vector<std::vector<T>> &a) {std::cerr << "{\n";for (const auto &v : a) {std::cerr << "\t";view(v);std::cerr << "\n";}std::cerr << "}";}template <class T, class U> void view(const std::vector<std::pair<T, U>> &a) {std::cerr << "{\n";for (const auto &p : a)std::cerr << "\t(" << p.first << ", " << p.second << ")\n";std::cerr << "}";}template <class T, class U> void view(const std::map<T, U> &m) {std::cerr << "{\n";for (const auto &p : m)std::cerr << "\t[" << p.first << "] : " << p.second << "\n";std::cerr << "}";}template <class T, class U> void view(const std::pair<T, U> &p) {std::cerr << "(" << p.first << ", " << p.second << ")";}template <class T> void view(const std::set<T> &s) {std::cerr << "{ ";for (auto &v : s) {view(v);std::cerr << ", ";}std::cerr << "\b\b }";}template <class T> void view(const T &e) { std::cerr << e; }} // namespace debugger#ifdef LOCALvoid debug_out() {}template <typename Head, typename... Tail> void debug_out(Head H, Tail... T) {debugger::view(H);std::cerr << ", ";debug_out(T...);}#define debug(...) \do { \std::cerr << __LINE__ << " [" << #__VA_ARGS__ << "] : ["; \debug_out(__VA_ARGS__); \std::cerr << "\b\b]\n"; \} while (false)#else#define debug(...) (void(0))#endiflong long power(long long x, long long n) {// O(logn)// https://algo-logic.info/calc-pow/#toc_id_1_2long long ret = 1;while (n > 0) {if (n & 1)ret *= x; // n の最下位bitが 1 ならば x^(2^i) をかけるx *= x;n >>= 1; // n を1bit 左にずらす}return ret;}// @brief nCr. O(r log n)。あるいは前処理 O(n), 本処理 O(1)で求められる modint// の bc を検討。int comb(int n, int r) {// https://www.geeksforgeeks.org/program-to-calculate-the-value-of-ncr-efficiently/if (n < r)return 0;// p holds the value of n*(n-1)*(n-2)...,// k holds the value of r*(r-1)...long long p = 1, k = 1;// C(n, r) == C(n, n-r),// choosing the smaller valueif (n - r < r)r = n - r;if (r != 0) {while (r) {p *= n;k *= r;// gcd of p, klong long m = __gcd(p, k);// dividing by gcd, to simplify// product division by their gcd// saves from the overflowp /= m;k /= m;n--;r--;}// k should be simplified to 1// as C(n, r) is a natural number// (denominator should be 1 ) .}elsep = 1;// if our approach is correct p = ans and k =1return p;}// MODvoid add(long long &a, long long b) {a += b;if (a >= MOD)a -= MOD;}template <class T> inline bool chmin(T &a, T b) {if (a > b) {a = b;return true;}return false;}template <class T> inline bool chmax(T &a, T b) {if (a < b) {a = b;return true;}return false;}// 負数も含む丸めlong long ceildiv(long long a, long long b) {if (b < 0)a = -a, b = -b;if (a >= 0)return (a + b - 1) / b;elsereturn a / b;}long long floordiv(long long a, long long b) {if (b < 0)a = -a, b = -b;if (a >= 0)return a / b;elsereturn (a - b + 1) / b;}long long floorsqrt(long long x) {assert(x >= 0);long long ok = 0;long long ng = 1;while (ng * ng <= x)ng <<= 1;while (ng - ok > 1) {long long mid = (ng + ok) >> 1;if (mid * mid <= x)ok = mid;elseng = mid;}return ok;}// @brief a^n mod modlong long modpower(long long a, long long n, long long mod) {long long res = 1;while (n > 0) {if (n & 1)res = res * a % mod;a = a * a % mod;n >>= 1;}return res;}// @brief s が c を含むかtemplate <class T> bool contain(const std::string &s, const T &c) {return s.find(c) != std::string::npos;}__attribute__((constructor)) void faster_io() {ios_base::sync_with_stdio(false);cin.tie(nullptr);cerr.tie(nullptr);}/* #endregion */using Graph = vector<vector<int>>;int ans = 0;template <typename T> class RBST {private:struct node {T val;node *left, *right;int st_size; // 部分木のサイズnode() {}node(T v) : val(v), left(nullptr), right(nullptr), st_size(1) {}~node() {delete left;delete right;}};node *root;using pnn = pair<node *, node *>;int size(node *t) { return t ? t->st_size : 0; }node *update(node *t) {node *l = t->left;node *r = t->right;t->st_size = size(l) + size(r) + 1;return t;}unsigned rnd() {static unsigned x = 123456789, y = 362436069, z = 521288629,w = 86675123;unsigned t = (x ^ (x << 11));x = y, y = z, z = w;return (w = (w ^ (w >> 19)) ^ (t ^ (t >> 8)));}node *merge(node *l, node *r) {if (!l || !r)return (!l) ? r : l;if (rnd() % (size(l) + size(r)) < (unsigned)size(l)) {l->right = merge(l->right, r);return update(l);} else {r->left = merge(l, r->left);return update(r);}}pnn split(node *t, int k) { //木のサイズが(k,n-k)となるように分割するif (!t)return pnn(nullptr, nullptr);if (k <= size(t->left)) {pnn s = split(t->left, k);t->left = s.second;return pnn(s.first, update(t));} else {pnn s = split(t->right, k - size(t->left) - 1);t->right = s.first;return pnn(update(t), s.second);}}int lower_bound(node *t, const T k) {if (!t)return 0;if (t->val < k) {return size(t->left) + lower_bound(t->right, k) + 1;} else {return lower_bound(t->left, k);}}void lower_value(node *t, const T k, T &res) {if (!t)return;if (t->val < k) {lower_value(t->right, k, res);} else {lower_value(t->left, k, res = t->val);}}int upper_bound(node *t, const T k) {if (!t)return 0;if (t->val <= k) {return size(t->left) + upper_bound(t->right, k) + 1;} else {return upper_bound(t->left, k);}}void upper_value(node *t, const T k, T &res) {if (!t)return;if (t->val <= k) {upper_value(t->right, k, res);} else {upper_value(t->left, k, res = t->val);}}T get(node *t, int k) {if (!t)assert(false);int s = size(t->left);if (s > k)return get(t->left, k);else if (s < k)return get(t->right, k - s - 1);elsereturn t->val;}node *insert(node *t, int k, node *u) {pnn s = split(t, k);return merge(merge(s.first, u), s.second);}pnn erase(node *t, int k) {pnn sr = split(t, k + 1);pnn sl = split(sr.first, k);return pnn(merge(sl.first, sr.second), sl.second);}public:RBST() : root(nullptr) {}// k以上の数の最小インデックスint lower_bound(const T k) { return lower_bound(root, k); }// k以上の最小の数T lower_value(const T k) {T res = numeric_limits<T>::max();lower_value(root, k, res);return res;}// kを超える数の最小インデックスint upper_bound(const T k) { return upper_bound(root, k); }// kを超える最小の数T upper_value(const T k) {T res = numeric_limits<T>::max();upper_value(root, k, res);return res;}//値valを挿入void insert(T val) { root = insert(root, upper_bound(val), new node(val)); }//値valを削除void erase(T val) {node *p;tie(root, p) = erase(root, lower_bound(val));p->left = p->right = nullptr;delete p;}// k番目の値を返すT get(int k) { return get(root, k); }void print() {int sz = size(root);REP(i, sz) cout << get(i) << " ";cout << "\n";}int size() { return size(root); }// a, b をマージする.a に全要素が入り,b が空になる.RBST<int> merge_tech(RBST<int> &a, RBST<int> &b) {if (a.size() < b.size()) {swap(a, b);}REP(i, b.size()) { a.insert(b.get(i)); }return a;}};// https://kopricky.github.io/code/BinarySearchTree/rbst_set.html/*signed main() {RBST<PII> rs;REP(ni, n) {rs.insert(MP(X[ni], ni));if (ni + 1 >= k)print(rs.get(k - 1).SS + 1);}return 0;}*/vector<RBST<int>> dp;RBST<int> rs;// 木上の探索// v: 現在探索中の頂点、p: v の親 (v が根のときは -1)void dfs(const Graph &G, int v, int p) {for (auto nv : G[v]) {if (nv == p)continue; // nv が親 p だったらダメdfs(G, nv, v); // v は新たに nv の親になります}dp[v].insert(v);for (auto c : G[v]) {if (c == p)continue;dp[v] = rs.merge_tech(dp[v], dp[c]);// REP(ni, SZ(dp[c])) { dp[v].insert(dp[c].get(ni)); }}// print("root", v);// dp[v].print();int idx = dp[v].upper_bound(v);ans += dp[v].size() - idx;}signed main() {// 頂点数 (木なので辺数は N-1 で確定)int n;cin >> n;// グラフ入力受取Graph G(n);for (int i = 0; i < n - 1; ++i) {int v;cin >> v;G[i + 1].push_back(v);G[v].push_back(i + 1);}debug(G);// exit(0);dp.resize(n);// 探索int root = 0; // 仮に頂点 0 を根とするdfs(G, root, -1);print(ans);}