【bzoj4311】向量 线段树按时间分治+凸包+三分

好像没什么好说的，记录每个点进出的时刻，每个点对应线段树上的O(logn)个节点，然后按时间分治就可以了。

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cmath>
#include<algorithm>
#include<iostream>
#include<vector>
#define maxn 200010

using namespace std;

struct yts1
{
long long x,y;
}p[maxn],st[maxn],tmp[maxn];

struct yts
{
int l,r;
vector<yts1> v;
}t[4*maxn];

struct yts2
{
long long x,y,ans;
}q[maxn];

int n,m,L[maxn],tot,top;

long long operator^ (yts1 a,yts1 b)
{
return a.x*b.y-a.y*b.x;
}

long long operator* (yts1 a,yts1 b)
{
return a.x*b.x+a.y*b.y;
}

bool operator< (yts1 a,yts1 b)
{
return a.x<b.x || (a.x==b.x && a.y<b.y);
}

yts1 operator-(yts1 a,yts1 b)
{
yts1 ans;
ans.x=a.x-b.x;ans.y=a.y-b.y;
return ans;
}

void build(int i,int l,int r)
{
t[i].l=l;t[i].r=r;
if (l==r) return;
int mid=(l+r)/2;
build(i*2,l,mid);build(i*2+1,mid+1,r);
}

void insert(int i,int l,int r,int x)
{
if (l<=t[i].l && t[i].r<=r)
{
t[i].v.push_back(p[x]);return;
}
int mid=(t[i].l+t[i].r)/2;
if (l<=mid) insert(i*2,l,r,x);
if (mid<r) insert(i*2+1,l,r,x);
}

void query(int i)
{
if (!top) return;
yts1 o;
o.x=q[i].x;o.y=q[i].y;
int l=1,r=top;
while (r-l>=3)
{
int mid=l+(r-l)/3,midmid=r-(r-l)/3;
if ((o*st[mid])>(o*st[midmid])) r=midmid; else l=mid;
}
for (int j=l;j<=r;j++) q[i].ans=max(q[i].ans,o*st[j]);
}

void solve(int x,int l,int r)
{
int mid=(l+r)/2;
if (l<r) solve(x*2,l,mid),solve(x*2+1,mid+1,r);
top=0;
sort(t[x].v.begin(),t[x].v.end());
for (int i=0;i<t[x].v.size();i++)
{
while (top>1 && ((t[x].v[i]-st[top])^(st[top]-st[top-1]))<=0) top--;
st[++top]=t[x].v[i];
}
for (int i=l;i<=r;i++)
if (q[i].x) query(i);
}

int main()
{
scanf("%d",&n);
build(1,1,n);
for (int i=1;i<=n;i++)
{
int opt;
long long x,y;
scanf("%d",&opt);
if (opt==1) {scanf("%lld%lld",&x,&y);p[++tot].x=x,p[tot].y=y,L[tot]=i;}
if (opt==2) {scanf("%lld",&x);insert(1,L[x],i,x);L[x]=0;}
if (opt==3) scanf("%lld%lld",&q[i].x,&q[i].y);
}
for (int i=1;i<=tot;i++) if (L[i]) insert(1,L[i],n,i);
solve(1,1,n);
for (int i=1;i<=n;i++)
if (q[i].x) printf("%lld\n",q[i].ans);
return 0;
}