CF 678F Lena and Queries

本文主要是介绍CF 678F Lena and Queries，希望对大家解决编程问题提供一定的参考价值，需要的开发者们随着小编来一起学习吧！

一个空集，有 $n(≤30w)$ 个操作：
1)集合加入对 $(a,b)(-10^9≤a,b≤10^9)$
2)移除 $i_{th}$ 操作加入的对
3)给出 $q(-10^9≤q≤10^9)$ ，对于集合中所有的(x,y)，求x·q+y的最大值。
PS:这里的集合应该允许同一元素出现多次，或者加入时间也包含在元素的信息中。
分治+凸壳
PS:有对 HDU 5390 tree 的线段树的一点改进。

#include <set>
#include <cmath>
#include <queue>
#include <stack>
#include <cstdio>
#include <bitset>
#include <cassert>
#include <cstring>
#include <complex>
#include <iostream>
#include <algorithm>
#define pi acos(-1)
#define inf (1<<30)
#define INF (1ll<<62)
#define y1 bflaisfnmasf
#define y2 fsafgmalg
#define tm afnsjkf
#define j1 sfakf
#define j2 fasndfkas
#define fi first
#define se second
#define CLR(x,f) memset(x,f,sizeof(x))
#define CPY(x,y) memcpy(x,y,sizeof(x))
#define prt(x) cout<<#x<<":"<<x<<" "
#define prtn(x) cout<<#x<<":"<<x<<endl
#define huh(x) printf("--------case(%d)--------\n",x)
#define travel(x) for(Edge *e=h[x];e;e=e->n)
#pragma comment(linker, "/STACK:1024000000,1024000000")
using namespace std;
const int M=3e5+4;template<class T>
void Max(T &x,T y){if(x<y)x=y;
}//nice!typedef long long ll;
typedef double doub;
struct Point{ll x,y;bool operator<(const Point &a)const{return x==a.x?y<a.y:x<a.x;}Point operator-(const Point &a)const{return (Point){x-a.x,y-a.y};}ll operator*(const Point &a)const{return x*a.x+a.y*y;}ll operator^(const Point &a)const{return x*a.y-a.x*y;}//cross(a,b)?
}po[M],s[M],g[M];
struct Pt{int l,r;Point p;
}p[M];
struct Qr{int t;ll x;
}q[M];
int conpoly(int n){int w=0;sort(s+1,s+n+1);for(int i=1;i<=n;i++){while(w>1&&(s[w]-s[w-1]^s[i]-s[w-1])>=0)w--;s[++w]=s[i];}return w;
}//仅保留凸包 
void calc(int n){for(int i=1;i<n;i++){g[i]=s[i+1]-s[i];g[i].y=-g[i].y;}
}
bool cmp(Point a,Point b){return a.y*b.x<b.y*a.x;
}
ll query(ll x,int n){if(n==1)return x*s[1].x+s[1].y;int p=lower_bound(g+1,g+n,(Point){1,x},cmp)-g;ll res=-INF;for(int i=-1;i<=1;i++)if(p+i>0&&p+i<=n)Max(res,x*s[p+i].x+s[p+i].y);//nice!//  ll tmp=-INF;
//  for(int i=1;i<=n;i++)
//      Max(tmp,x*s[i].x+s[i].y);
//  assert(tmp==res);return res;
}
ll ans[M];
void solve(int n,int m,int l,int r){if(!n||!m||l>r)return;int w=0;for(int i=0;i<n;i++)if(p[i].l<=l&&r<=p[i].r)s[++w]=p[i].p;if(w>0){w=conpoly(w);calc(w);for(int i=0;i<m;i++)Max(ans[q[i].t],query(q[i].x,w));}if(l==r)return;int mid=l+r>>1,w0,w1;w0=w1=0;for(int i=0;i<n;i++)if(p[i].l<=mid&&!(p[i].l<=l&&r<=p[i].r))swap(p[w0++],p[i]);for(int i=0;i<m;i++)if(q[i].t<=mid)swap(q[w1++],q[i]);solve(w0,w1,l,mid);w0=w1=0;for(int i=0;i<n;i++)if(p[i].r>mid&&!(p[i].l<=l&&r<=p[i].r))swap(p[w0++],p[i]);for(int i=0;i<m;i++)if(q[i].t>mid)swap(q[w1++],q[i]);solve(w0,w1,mid+1,r);
}
int en[M],type[M];
int main(){int n;scanf("%d",&n);int w1,w2;w1=w2=0;for(int u,v,i=0;i<n;i++){scanf("%d",&type[i]);//1)Add a pair (a,b) to the set.//2)Remove a pair added in the query number i. All queries are numbered with integers from 1 to n.//3)For a given integer q find the maximal value x·q?+?y over all pairs (x,y) from the set.switch(type[i]){case 1:scanf("%d%d",&u,&v);po[i]=(Point){u,v};en[i]=n-1;break;case 2:scanf("%d",&u);en[u-1]=i;break;case 3:scanf("%d",&u);q[w2++]=(Qr){i,u};ans[i]=-INF;break;default:break;//}}for(int i=0;i<n;i++)if(type[i]==1)p[w1++]=(Pt){i,en[i],po[i]};solve(w1,w2,0,n-1);for(int i=0;i<n;i++){if(type[i]==3){if(ans[i]==-INF)puts("EMPTY SET");else printf("%I64d\n",ans[i]);}}return 0;
}