【bzoj2300】【Luogu P2521】 [HAOI2011]防线修建 动态凸包，平衡树，Set

一句话题意：给你一个凸包，每次可以插入一个点或者询问周长。

动态凸包裸题嘛，用(Set)实现。最初每个点坐标做乘三处理，便于取初始三角形的重心作为凸包判定原点。

#include <bits/stdc++.h>
using namespace std;

const int N = 500000 + 5;
const double eps = 1e-8;

int n, m, x, y, q, px[N], py[N], ban[N];

struct Point {
	int x, y; double ang;
}o;

Point operator - (Point lhs, Point rhs) {
	return (Point) {lhs.x - rhs.x, lhs.y - rhs.y};
}

double disPP (Point lhs, Point rhs) {
	return hypot (abs (lhs.x - rhs.x), abs (lhs.y - rhs.y));
}

bool operator < (Point lhs, Point rhs) {
	if (fabs (lhs.ang - rhs.ang) > eps) {
		return lhs.ang < rhs.ang;
	} else {
		return disPP (lhs, o) < disPP (rhs, o);
	}
}

int Cross (Point lhs, Point rhs) {
	return lhs.x * rhs.y - lhs.y * rhs.x;
}

struct Query {int opt, x;}qry[N];

set <Point> s;

typedef set <Point> :: iterator Iter;

double cur_ans = 0;

stack <double> ans;

Iter Prev (Iter it) {
	return it == s.begin () ? (--s.end ()) : (--it);
}

Iter Next (Iter it) {
	return (++it) == s.end () ? s.begin () : it; 
}

void Insert (Point P) {
	Iter nxt = s.lower_bound (P); 
	if (nxt == s.end ()) {
		nxt = s.begin ();
	}
	Iter pre = Prev (nxt);
	if (Cross (P - *pre, *nxt - P) <= 0) {
		return; // 已经在凸包里面 
	}
	s.insert (P); 
	Iter i, j, cur = s.find (P);
	i = Prev (cur), j = Prev (i);
	pre = Prev (cur), nxt = Next (cur);
	cur_ans += disPP (*cur, *pre);
	cur_ans += disPP (*cur, *nxt);
	cur_ans -= disPP (*pre, *nxt);
	while (Cross (*i - *j, *cur - *i) <= 0) {
		cur_ans -= disPP (*cur, *i);
		cur_ans -= disPP (*i, *j);
		cur_ans += disPP (*cur, *j);
		s.erase (i); i = j; j = Prev (j);
	} 
	i = Next (cur), j = Next (i);
	while (Cross (*i - *cur, *j - *i) <= 0) {
		cur_ans -= disPP (*cur, *i);
		cur_ans -= disPP (*i, *j);
		cur_ans += disPP (*cur, *j);
		s.erase (i); i = j; j = Next (j);
	}
}

int main () {
//	freopen ("data.in", "r", stdin);
	cin >> n >> x >> y >> m;
	x *= 3, y *= 3, n *= 3;
	for (int i = 1; i <= m; ++i) {
		cin >> px[i] >> py[i];
		px[i] *= 3; py[i] *= 3;
	}
	cin >> q;
	for (int i = 1; i <= q; ++i) {
		cin >> qry[i].opt;
		if (qry[i].opt == 1) {
			cin >> qry[i].x;
			ban[qry[i].x] = true;
		}	
	}
	o = (Point) {(n + x) / 3, y / 3, 0};
	Point P1 = (Point) {0, 0, atan2 (0 - o.y, 0 - o.x)};
	Point P2 = (Point) {n, 0, atan2 (0 - o.y, n - o.x)};
	Point P3 = (Point) {x, y, atan2 (y - o.y, x - o.x)};
	s.insert (P1);
	s.insert (P2); cur_ans += disPP (P2, P3);
	s.insert (P3); cur_ans += disPP (P1, P3);
	for (int i = 1; i <= m; ++i) {
		if (!ban[i]) {
			Insert ((Point) {px[i], py[i], atan2 (py[i] - o.y, px[i] - o.x)});
		}
	}
	for (int i = q; i >= 1; --i) {
		if (qry[i].opt == 1) {
			int t = qry[i].x;
			Insert ((Point) {px[t], py[t], atan2 (py[t] - o.y, px[t] - o.x)});
		} else {
			ans.push (cur_ans);
		}
	}
	while (!ans.empty ()) {
		cout << fixed << setprecision (2) << ans.top () / 3.0 << endl;
		ans.pop ();
	}
}