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 ();}
}