結果
問題 | No.913 木の燃やし方 |
ユーザー |
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提出日時 | 2020-02-23 17:59:27 |
言語 | C++14 (gcc 13.3.0 + boost 1.87.0) |
結果 |
AC
|
実行時間 | 710 ms / 3,000 ms |
コード長 | 4,719 bytes |
コンパイル時間 | 1,383 ms |
コンパイル使用メモリ | 133,336 KB |
実行使用メモリ | 18,864 KB |
最終ジャッジ日時 | 2024-10-10 06:45:31 |
合計ジャッジ時間 | 21,257 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 * 34 |
ソースコード
# include <iostream># include <algorithm>#include <array># include <cassert>#include <cctype>#include <climits>#include <numeric># include <vector># include <string># include <set># include <map># include <cmath># include <iomanip># include <functional># include <tuple># include <utility># include <stack># include <queue># include <list># include <bitset># include <complex># include <chrono># include <random># include <limits.h># include <unordered_map># include <unordered_set># include <deque># include <cstdio># include <cstring>#include <stdio.h>#include<time.h>#include <stdlib.h>#include <cstdint>#include <cfenv>#include<fstream>//#include <bits/stdc++.h>using namespace std;using LL = long long;using ULL = unsigned long long;long long MOD = 1000000000 + 7;// 998244353;// ;constexpr long long INF = numeric_limits<LL>::max() / 2;const double PI = acos(-1);#define fir first#define sec second#define thi third#define debug(x) cerr<<#x<<": "<<x<<'\n'typedef pair<LL, LL> Pll;typedef pair<LL, pair<LL, LL>> Ppll;typedef pair<LL, pair<LL, bitset<100001>>> Pbll;typedef pair<LL, pair<LL, vector<LL>>> Pvll;typedef pair<LL, LL> Vec2;struct Tll { LL first, second, third; };struct Fll { LL first, second, third, fourd; };typedef pair<LL, Tll> Ptll;#define rep(i,rept) for(LL i=0;i<rept;i++)#define Rrep(i,mf) for(LL i=mf-1;i>=0;i--)void YN(bool f) {if (f)cout << "YES" << endl;elsecout << "NO" << endl;}void yn(bool f) {if (f)cout << "Yes" << endl;elsecout << "No" << endl;}struct Edge { LL to, cost; };struct edge { LL from, to, cost; };vector<vector<LL>>g, rev;vector<edge>edges;vector<LL>v;map<LL, LL>ma;set<LL>st;LL h, w, n, m, k, t, s, p, q, last, cnt, sum[210000], ans[210000], a[510000], b[510000], dp[210000];string str, ss;bool f;char c;template<typename T>class ConvexHullTrick {private:// 直線群(配列)std::vector<std::pair<T, T>> lines;// 最小値(最大値)を求めるxが単調であるかbool isMonotonicX;// 最小/最大を判断する関数std::function<bool(T l, T r)> comp;public:// コンストラクタ ( クエリが単調であった場合はflag = trueとする )ConvexHullTrick(bool flagX = false, std::function<bool(T l, T r)> compFunc = [](T l, T r) {return l >= r; }):isMonotonicX(flagX), comp(compFunc) {//lines.emplace_back(0, 0);//これいるかどうかはわからない};void clear() {lines.clear();}// 直線l1, l2, l3のうちl2が不必要であるかどうかbool check(std::pair<T, T> l1, std::pair<T, T> l2, std::pair<T, T> l3) {if (l1 < l3) std::swap(l1, l3);return (l3.second - l2.second) * (l2.first - l1.first) >= (l2.second - l1.second) * (l3.first - l2.first);//return comp((l3.second - l2.second) * (l2.first - l1.first), (l2.second - l1.second) * (l3.first - l2.first));}// 直線y=ax+bを追加するvoid add(T a, T b) {std::pair<T, T> line(a, b);while (lines.size() >= 2 && check(*(lines.end() - 2), lines.back(), line))lines.pop_back();lines.emplace_back(line);}// i番目の直線f_i(x)に対するxの時の値を返すT f(int i, T x) {return lines[i].first * x + lines[i].second;}// i番目の直線f_i(x)に対するxの時の値を返すT f(std::pair<T, T> line, T x) {return line.first * x + line.second;}// 直線群の中でxの時に最小(最大)となる値を返すT get(T x) {// 最小値(最大値)クエリにおけるxが単調if (isMonotonicX) {static int head = 0;while (lines.size() - head >= 2 && comp(f(head, x), f(head + 1, x)))++head;return f(head, x);}else {int low = -1, high = lines.size() - 1;while (high - low > 1) {int mid = (high + low) / 2;(comp(f(mid, x), f(mid + 1, x)) ? low : high) = mid;}return f(high, x);}}};//[L,R)LL chmin[210000];void dfs(LL L,LL R) {if (R-L<=2) {return;}ConvexHullTrick<LL> cht;LL mid = (L + R) / 2;for (LL i =R; i > mid; i--) {cht.add(-2 * i, i * i + sum[i]);}chmin[L] = INF;for (LL i = L; i <= mid;i++) {chmin[i] = min(chmin[i], cht.get(i) + i * i - sum[i]);chmin[i + 1] = chmin[i];ans[i] = min(ans[i], chmin[i]);}cht.clear();for (LL i = L; i <=mid; i++) {cht.add(-2 * i, i * i - sum[i]);}chmin[R - 1] = INF;for (LL i = R; i > mid; i--) {chmin[i - 1] = min(cht.get(i) + i * i + sum[i], chmin[i - 1]);chmin[i - 2] = chmin[i - 1];ans[i - 1] = min(ans[i - 1], chmin[i - 1]);}dfs(L, (L + R) / 2);dfs((L + R) / 2, R);}int main() {cin >> n;rep(i, n) {cin >> a[i];sum[i + 1] = sum[i] + a[i];ans[i] = a[i] + 1;}dfs(0,n);rep(i, n)cout << ans[i] << endl;return 0;}