Submission #826292

#	Time	Username	Problem	Language	Result	Execution time	Memory
826292		NK_	Building Bridges (CEOI17_building)	C++17	100 / 100	41 ms	8128 KiB

building

// Success consists of going from failure to failure without loss of enthusiasm
#include <bits/stdc++.h>

using namespace std;

#define nl '\n'
#define pb push_back 
#define mp make_pair
#define f first
#define s second
#define sz(x) int(x.size())

template<class T> using V = vector<T>;
using vi = V<int>;
using ll = long long;
using pl = pair<ll, ll>;
using vl = V<ll>;
// using vpi = V<pi>;

const ll INFL = ll(1e18) + 1008;
const int nax = int(1e6);


struct line {
	ll m, b; 
	line() { m = 0, b = INFL; }
	line(ll m, ll b) : m(m), b(b) { }
	ll get(ll x) { return m * x + b; }
};

struct LiChao {
	struct node {
		node *l, *r;
		line x;
		node(line x) : x(x) { l = r = 0; }
	};

	node *t = new node(line());

	void upd(line x, node *&v, int l = 0, int r = nax) {
		if (!v) { v = new node(x); return; }
		// cout << l << " <-> " << r << endl;
		int m = (l + r) / 2;
		bool lf = x.get(l) < v->x.get(l);
		bool md = x.get(m) < v->x.get(m);
		if (md) swap(x, v->x);
		if (l == r) return;
		if (lf != md) upd(x, v->l, l, m);
		else upd(x, v->r, m+1, r);
	}

	ll get(int x, node *&v, int l = 0, int r = nax) {
		if (!v) return INFL;
		// cout << l << " < - > " << r << endl;
		int m = (l + r) / 2;
		if (l == r) return v->x.get(x);
		if (x < m) return min(v->x.get(x), get(x, v->l, l, m));
		else return min(v->x.get(x), get(x, v->r, m+1, r));
	}

	void upd(line x) { return upd(x, t); }
	ll get(int x) { return get(x, t); }
};

int main() {
	cin.tie(0)->sync_with_stdio(0);

	int N; cin >> N;
	vi A(N); for(auto& x : A) cin >> x;
	vi W(N); for(auto& x : W) cin >> x;

	if (A.front() > A.back()) {
		reverse(begin(A), end(A));
		reverse(begin(W), end(W));
	}

	vl P = {0}; for(auto& x : W) P.pb(P.back() + x);

	auto sq = [&](int x) { return x * 1LL * x; };

	vl dp(N, INFL); dp[0] = 0;
	LiChao T; T.upd(line(-2 * A[0], dp[0] + sq(A[0]) - P[1]));

	for(int i = 1; i < N; i++) {
		dp[i] = T.get(A[i]) + sq(A[i]) + P[i];
		T.upd(line(-2 * A[i], dp[i] + sq(A[i]) - P[i + 1]));
	}

	cout << dp.back() << nl;
    return 0;
}			

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