제출 #529111

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
529111		Graphter	Aliens (IOI16_aliens)	C++17	100 / 100	652 ms	5028 KiB

aliens

#include <bits/stdc++.h>
#define debug(x) cout << #x << " = " << x << endl
#define REP(i, n) for (Long i = 0; i < (Long)n; i++)
using namespace std;

typedef long long Long;

//https://contest.yandex.com/ioi/contest/2830/problems/F/

const Long INF = 1e18;

struct Line {
	mutable Long m, b;
	//mx + b, intersect with next line at r (rounded down)
	Line() {}
	Line(Long m, Long b): m(m), b(b){}
	Long getVal(Long x) {return m * x + b;}
};

struct CHT{
	deque<Line> envelope;

	Long div(Long a, Long b) { //floored division
		//CAREFUL ! this won't produced the right convex envelope
		//but the maxY function will still work for integers
		//if you need the correct convex envelope, use double division
		//or multiplication in "bad" function
		assert(b != 0);
		return a / b - ((a ^ b) < 0 && a % b); 
	}
	
	Long intersect(Line l1, Line l2) {
		return div(l1.b - l2.b, l2.m - l1.m);
	}
	
	bool bad(Line l1, Line l2, Line l3) {
		//tells if l2 is bad an can be eliminated by l3
		return intersect(l2 , l3) <= intersect(l1, l2);
	}

	void addLine(Line l3) { 
		l3.m *= -1;
		l3.b *= -1;
		if (envelope.empty()) {
			envelope.push_back(l3);
			return;
		}
		if (l3.m == envelope.back().m) {
			if (l3.b > envelope.back().b) envelope.pop_back();
			else return;
		}
		if (envelope.size() <= 1) {
			envelope.push_back(l3);
			return;
		}
		Long sz = envelope.size();
		Line l1 = envelope[sz - 2];
		Line l2 = envelope[sz - 1];
		while (bad(l1, l2, l3)) {
			envelope.pop_back();
			if (envelope.size() == 1) break;
			sz = envelope.size();
			l1 = envelope[sz - 2];
			l2 = envelope[sz - 1];
		}
		envelope.push_back(l3);
	}
	
	Long minY(Long x) { //O(log n)
		assert(!envelope.empty());
		while (envelope.size() >= 2 && envelope[0].getVal(x) < envelope[1].getVal(x)) {
			envelope.pop_front();
		}
		return -envelope[0].getVal(x);
	}
};

struct Range {
	int l, r;
	Range(int l, int r): l(l), r(r){}
	bool operator <(const Range &other) {
		return r < other.r;
	}
	bool contains(Range &other) {
		return l <= other.l && other.r <= r;
	}
};

void removeContained(vector<Range> &v) {
	int n = v.size();
	vector<Range> ans = {v.back()};
	for (int i = n - 2; i >= 0; i--) {
		if (ans.back().contains(v[i])) continue;
		ans.push_back(v[i]);
	}
	reverse(ans.begin(), ans.end());
	v = ans;
}

Long square(Long x) {
	return x * x;
}

vector<int> L, R;
Long cost(int l, int r) {
	Long ans = square(R[r] - L[l]);
	if (l > 0) ans -= square(max(0, R[l - 1] - L[l]));
	return ans;
}

Long getBest(Long lambda, int k) {
	int n = L.size();
	CHT cht;
	vector<Long> dp(n);
	auto getLine = [&](int i) {
		return Line(-2 * L[i + 1], 
		dp[i] + square(L[i + 1]) - square(max(0, R[i] - L[i + 1])));
	};
	
	cht.addLine(Line(-2 * L[0], square(L[0])));
	for (int i = 0; i < n; i++) {
		dp[i] = cht.minY(R[i]);
		dp[i] += square(R[i]) - lambda;
		if (i + 1 < n) cht.addLine(getLine(i));
	}
	return dp[n - 1] + lambda * k;
}

const Long INF2 = 1e13;

Long minCost(int k) { //O(log n)
	Long low = -INF2;
	Long high = INF2;
	while (high - low > 2) {
		Long m1 = low + (high - low) / 3;
		Long m2 = high - (high - low) / 3;
		if (getBest(m1, k) < getBest(m2, k)) low = m1;
		else high = m2;
	}
	Long maxi = getBest(low, k); 
	for (Long i = low + 1; i <= high; i++) {
		maxi = max(maxi, getBest(i, k));
	}
	return maxi;
}

Long take_photos(int n, int m, int K, vector<int> x, vector<int> y) {
	vector<Range> v;
	REP(i, n) {
		int l = min(x[i], y[i]);
		int r = max(x[i], y[i]);
		v.push_back({l, r});
	}
	sort(v.begin(), v.end());
	removeContained(v);
	n = v.size();
	L = R = vector<int>(n);
	REP(i, n) {
		v[i].r++;
		L[i] = v[i].l;
		R[i] = v[i].r;
	}
	K = min(K, n);
	return minCost(K);
}
/*
5 8 5
1 5
3 4
2 6
1 4
3 5 
ans = 34

7 30 5
1 5
3 4
2 6
1 4
3 5 
9 15
14 22
ans = 160
*/
/*int main() {
	ios_base::sync_with_stdio(false);
	cin.tie(0);
	
	int n, m, k;
	cin >> n >> m >> k;
	vector<int> x(n);
	vector<int> y(n);
	REP(i, n) {
		cin >> x[i] >> y[i];
	}
	cout << take_photos(n, m, k, x, y) << "\n";
	
	return 0;
}*/

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