Submission #135863

#	Time	Username	Problem	Language	Result	Execution time	Memory
135863		Kastanda	Aliens (IOI16_aliens)	C++11	100 / 100	800 ms	17196 KiB

aliens

// ItnoE
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef double ld;
struct CHT
{
	typedef pair < ld , ld > Line;
	vector < pair < ld , pair < Line , ll > > > A;
	ld INF = (ld)1e18;
	inline void Clear()
	{
		A.clear();
	}
	inline void Add(Line X, ll id)
	{
		while (A.size() && Intersection(A.back().second.first, X) <= A.back().first)
			A.pop_back();
		if (A.size())
			A.push_back({Intersection(A.back().second.first, X), {X, id}});
		else
			A.push_back({-INF, {X, id}});
	}
	inline pair < ld , ll > GetMax(ld X)
	{
		int lb = upper_bound(A.begin(), A.end(), pair < ld , pair < Line , ll > > {X, {{INF, INF}, -1}}) - A.begin() - 1;
		return (make_pair(A[lb].second.first.first * X + A[lb].second.first.second, A[lb].second.second));
	}
	inline ld Intersection(Line X, Line Y)
	{
		if (X.first == Y.first && X.second <= Y.second)
			return (-INF);
		if (X.first == Y.first)
			return (INF);
		return ((X.second - Y.second) / (Y.first - X.first));// + ((X.second - Y.second) % (Y.first - X.first) > 0);
	}
};
const int N = 100005, MXM = 1e6 + 10;
int n, m, A[N], B[N], F[N], MN[MXM];
pair < ld , ll > dp[N];
inline ld SQR(ld a) {return (a * a);}
inline pair < ld , ll > Solve(ld md)
{
	CHT C;
	dp[0] = {0, 0};
	for (int i = 1; i <= n; i ++)
	{
		ld vl = dp[i - 1].first + SQR(F[i]);
		if (i > 1 && B[i - 1] > F[i])
			vl -= SQR(B[i - 1] - F[i]);
        C.Add({F[i], - vl}, i);
		auto tt = C.GetMax(2 * B[i]);
		tt.first *= -1;
		dp[i].first = tt.first + SQR(B[i]) + md;
		dp[i].second = dp[tt.second - 1].second + 1;
	}
	return (dp[n]);
}
int64_t take_photos(int q, int mm, int k, vector < int > RG, vector < int > CG)
{
	m = mm;
	memset(MN, 63, sizeof(MN));
	for (int i = 0; i < q; i ++)
	{
		if (RG[i] > CG[i])
			swap(RG[i], CG[i]);
		MN[CG[i]] = min(MN[CG[i]], RG[i]);
	}
	for (int i = 0; i < m; i ++)
		if (MN[i] <= i)
		{
			int b = i, a = i - MN[i] + 1;
			while (n && B[n] - A[n] >= b - a)
				n --;
			n ++; B[n] = b; A[n] = a;
			F[n] = B[n] - A[n];
		}

	ld le = 0, ri = (ll)m * m + 1, md;
	k = min(k, n);
	for (int _ = 0; _ <= 70; _ ++)
	{
		md = (le + ri) / 2;
		auto X = Solve(md);
		if (X.second <= k)
			ri = md;
		else
			le = md;
	}
	ld res = Solve(ri).first - ri * k;
	ll ret = floor(res + 0.12);
	return (ret);
}

• Subtask 1: 4 / 4
• Subtask 2: 12 / 12
• Subtask 3: 9 / 9
• Subtask 4: 16 / 16
• Subtask 5: 19 / 19
• Subtask 6: 40 / 40