제출 #1117890

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
1117890		vjudge1	Building Bridges (CEOI17_building)	C++17	100 / 100	46 ms	22088 KiB

building

//#pragma GCC optimize("O3,unroll-loops")
//#pragma GCC target("avx2,bmi,bmi2,lzcnt,popcnt")
#include<bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#define ll long long
#define F first
#define S second
#define ull unsigned long long
#define db double
#define ldb long double
#define pb push_back
#define pf push_front
#define ppb pop_back
#define ppf pop_front
#define yes cout << "YES\n"
#define no cout << "NO\n"
#define ordered_set tree < ll, null_type, less < ll > , rb_tree_tag, tree_order_statistics_node_update >
#define all(x) x.begin(), x.end()

const int mod = 1e9 + 7;
const int N = 500001;

using namespace std;
using namespace __gnu_pbds;

ll n, m, a, b, c, p[N], f[N], dp[N], pr[N];

struct Line {
	mutable ll k, m, p;
	bool operator<(const Line& o) const { return k < o.k; }
	bool operator<(ll x) const { return p < x; }
};

struct LineContainer : multiset<Line, less<>> {
	// (for doubles, use inf = 1/.0, div(a,b) = a/b)
	static const ll inf = LLONG_MAX;
	ll div(ll a, ll b) { // floored division
		return a / b - ((a ^ b) < 0 && a % b); }
	bool isect(iterator x, iterator y) {
		if (y == end()) return x->p = inf, 0;
		if (x->k == y->k) x->p = x->m > y->m ? inf : -inf;
		else x->p = div(y->m - x->m, x->k - y->k);
		return x->p >= y->p;
	}
	void add(ll k, ll m) {
		auto z = insert({k, m, 0}), y = z++, x = y;
		while (isect(y, z)) z = erase(z);
		if (x != begin() && isect(--x, y)) isect(x, y = erase(y));
		while ((y = x) != begin() && (--x)->p >= y->p)
			isect(x, erase(y));
	}
	ll query(ll x) {
		assert(!empty());
		auto l = *lower_bound(x);
		return l.k * x + l.m;
	}
};

signed main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
	cin >> n;
	for (int i = 1; i <= n; i++){
		cin >> p[i];
	}
	for (int i = 1; i <= n; i++){
		cin >> f[i];
		pr[i] = f[i] + pr[i - 1];
	}
	LineContainer cht;
	dp[1] = 0;
	cht.add(p[1], -(p[1] * p[1] + dp[1] - pr[1]));
//	cout << -(dp[1] - pr[1] + p[1] * p[1]) << '\n';
	for (int i = 2; i <= n; i++){
		dp[i] = (p[1] - p[i]) * (p[1] - p[i]) + (pr[i - 1] - pr[1]); 
		dp[i] = min(p[i] * p[i] - cht.query(2 * p[i]) + pr[i - 1] , dp[i]);
		cht.add (p[i], -(p[i] * p[i] + dp[i] - pr[i]));
//		cout << dp[i] << ' ';
	}
	cout << dp[n];
}

• Subtask 1: 30 / 30
• Subtask 2: 30 / 30
• Subtask 3: 40 / 40