# Bzoj 4311: Vector (divided by time + line tree)

Source: Internet
Author: User

Offline, then divided by time, each vector has a time of occurrence [L, R], directly inserted into the time segment tree (a vector only affects the O (LOGN) Order of magnitude of the segment tree node). Make a convex hull and then two points on each node in the segment tree. Time complexity O (nlog^2n)

---------------------------------------------------------------------------

#include <cstdio>#include <cctype>#include <cstring>#include <algorithm>using namespace std;typedef long Long ll;#define V (x) v[p[x]]#define C (x) v[c[x]]#define Q (x) q[_q[x]]#define K (A, B) ((double) (A.Y-B.Y)/(a.x-b.x))const int MAXN = 200009;int N, T, qn, vn, Pn, CN;int P[MAXN], _Q[MAXN], C[MAXN];Double LK[MAXN], RK[MAXN];ll ANS[MAXN];int buf[20];inline int getint () {char C = getchar ();For (;!isdigit (c); c = GetChar ());int ret = 0;For (; IsDigit (c); c = GetChar ())ret = ret * + C-' 0 ';return ret;}inline void Putint (ll x) {if (!x) {puts ("0");} else {int n = 0;For (; x; x/=) buf[n++] = x%;while (n--) Putchar (buf[n] + ' 0 ');puts ("");}}struct Q {int p, x, y;} Q[MAXN]; struct V {int x, Y, L, R;} V[MAXN];struct L {int p;l* NXT;} LPOOL[MAXN *, *lpt = Lpool;inline void Addl (l*&t) {lpt->p = T;lpt->nxt = t;t = lpt++;}struct Node {Node *LC, *RC;l* v;} POOL[MAXN << 1], *pt = Pool, *root;void Modify (node* t, int l, int r) {if (v[t].l <= l && R <= v[t].r) {Addl (t->v);} else {int m = (L + r) >> 1;if (v[t].l <= m) Modify (T->LC, L, m);if (M < V[T].R) Modify (T->RC, M + 1, R);}}void Build (node* t, int l, int r) {if (l! = r) {int m = (L + r) >> 1;Build (T->LC = pt++, L, m);Build (T->RC = pt++, M + 1, R);}}bool CMP (const int &L, const int &r) {return v[l].x < v[r].x | | (v[l].x = = v[r].x && v[l].y < V[R].Y);} void Solve (node* t, int l, int r) {if (l! = r) {int m = (L + r) >> 1;Solve (T->LC, L, m);Solve (T->RC, M + 1, R);}PN = cn = 0;For (l* o = t->v; o; o = o->nxt) p[pn++] = o->p;if (!PN) return;sort (p, p + pn, CMP);c[cn++] = p[0];for (int i = 1; i < PN; i++) {while (CN && V (i). x = = C (cn-1). x) cn--;while (cn > 1 && K (V (i), C (cn-1)) > K (c (cn-1), C (cn-2))) cn--;c[cn++] = p[i];}lk[0] = 1e30, rk[cn-1] = -1e30;for (int i = 1; I < CN; i++)Lk[i] = rk[i-1] = K (c (i), C (i-1));for (int i = l; I <= R; i++) if (~_q[i]) {int _l = 0, _r = cn-1;Double k = -1.0 * Q (i). x/q (i). Y;While (_l <= _r) {int m = (_l + _r) >> 1;Ans[_q[i]] = max (Ans[_q[i]), LL (C (M). x) * Q (i). x + ll (c (m). Y) * Q (i). y);if (K < lk[m] && Rk[m] < k) {Ans[_q[i]] = max (Ans[_q[i]), LL (C (M). x) * Q (i). x + ll (c (m). Y) * Q (i). y);Break ;}(K < lk[m] && K < rk[m])? _l = m + 1: _r = m-1;}}}void Work () {Build (Root = pt++, 1, N);For (T = 0; T < VN; t++) Modify (Root, 1, N);pn = 0;Solve (Root, 1, N);for (int i = 0; i < qn; i++) Putint (Ans[i]);}void Init () {N = Getint ();qn = vn = 0;memset (_q,-1, sizeof _q);for (int i = 1; I <= N; i++) {int t = getint ();if (t = = 3) {_q[i] = qn;q[qn].x = Getint (), q[qn].y = Getint ();Q[QN++].P = i;} else if (t = = 1) {v[vn].x = Getint (), v[vn].y = Getint ();V[VN].L = i, V[VN++].R = N;} elsev[getint ()-1].R = i;}memset (ans, 0, sizeof ans);}Init ();Work ();return 0;}

---------------------------------------------------------------------------

4311: Vector time limit: Sec Memory limit: MB
Submit: Solved: 26
[Submit] [Status] [Discuss] Description you want to maintain a vector collection that supports the following: 1. Insert a vector (x, y) 2. Delete the inserted vector 3. Query the maximum value of the current collection and (x, y) dot product. If current is an empty set output 0

Input the first line to enter an integer n, indicating the number of operations next n lines, each line first an integer t to denote the type, if t=1, the input vector (x, y), if t=2, the input ID indicates the deletion of the ID vector, otherwise input (x, y), the query and vector (x, y) dot product maximum is what. Ensure that a vector is deleted only once, without deleting the vector that has not been inserted

Output for each t=3 query, export an answer

Sample Input5
1 3 3
1 1 4
3 3 3
2 1
3 3 3Sample Output18
theHINT

n<=200000 1<=x,y<=10^6

Source

Bzoj 4311: Vector (divided by time + line tree)

Related Keywords:

The content source of this page is from Internet, which doesn't represent Alibaba Cloud's opinion; products and services mentioned on that page don't have any relationship with Alibaba Cloud. If the content of the page makes you feel confusing, please write us an email, we will handle the problem within 5 days after receiving your email.

If you find any instances of plagiarism from the community, please send an email to: info-contact@alibabacloud.com and provide relevant evidence. A staff member will contact you within 5 working days.

## A Free Trial That Lets You Build Big!

Start building with 50+ products and up to 12 months usage for Elastic Compute Service

• #### Sales Support

1 on 1 presale consultation

• #### After-Sales Support

24/7 Technical Support 6 Free Tickets per Quarter Faster Response

• Alibaba Cloud offers highly flexible support services tailored to meet your exact needs.