BZOJ4311 向量（线段树分治+三分）

　　由点积的几何意义（即投影）可以发现答案一定在凸壳上，并且投影的变化是一个单峰函数，可以三分。现在需要处理的只有删除操作，线段树分治即可。

#include<iostream>

#include<cstdio>

#include<cmath>

#include<cstdlib>

#include<cstring>

#include<algorithm>

#include<vector>

using namespace std;

int read()

{

    int x=,f=;char c=getchar();

    while (c<''||c>'') {if (c=='-') f=-;c=getchar();}

    while (c>=''&&c<='') x=(x<<)+(x<<)+(c^),c=getchar();

    return x*f;

}

#define N 200010

#define ll long long

int n,m,t;

int L[N<<],R[N<<],size[N<<];

struct point

{

    int x,y,s;

    ll operator *(const point&a) const

    {

        return 1ll*x*a.x+1ll*y*a.y;

    }

    bool operator <(const point&a) const

    {

        return x<a.x;

    }

}a[N],q[N];

bool check(point a,point b,point c)

{

    return 1ll*(a.y-b.y)*(c.x-b.x)>1ll*(c.y-b.y)*(a.x-b.x);

}

vector<point> tree[N<<];

void build(int k,int l,int r)

{

    L[k]=l,R[k]=r;

    if (l==r) return;

    int mid=l+r>>;

    build(k<<,l,mid);

    build(k<<|,mid+,r);

}

void ins(int k,int l,int r,point x)

{

    if (L[k]==l&&R[k]==r) {tree[k].push_back(x);return;}

    int mid=L[k]+R[k]>>;

    if (r<=mid) ins(k<<,l,r,x);

    else if (l>mid) ins(k<<|,l,r,x);

    else ins(k<<,l,mid,x),ins(k<<|,mid+,r,x);

}

void make(int k)

{

    sort(tree[k].begin(),tree[k].end());

    int s=tree[k].size(),t=-;

    for (int i=;i<s;i++)

    {

        while (t>=&&tree[k][i].y>tree[k][t].y) t--;

        while (t>&&check(tree[k][t-],tree[k][t],tree[k][i])) t--;

        tree[k][++t]=tree[k][i];

    }

    size[k]=t;

    if (L[k]==R[k]) return;

    make(k<<),make(k<<|);

}

ll calc(int k,point x)

{

    if (size[k]==-) return ;

    int l=,r=size[k];

    while (l+<r)

    {

        int mid1=l+(r-l)/,mid2=r-(r-l)/;

        if (x*tree[k][mid1]>x*tree[k][mid2]) r=mid2;

        else l=mid1;

    }

    ll ans=;

    for (int i=l;i<=r;i++) ans=max(ans,x*tree[k][i]);

    return ans;

}

ll query(int k,int p,point x)

{

    ll t=calc(k,x);

    if (L[k]==R[k]) return t;

    int mid=L[k]+R[k]>>;

    if (p<=mid) return max(t,query(k<<,p,x));

    else return max(t,query(k<<|,p,x));

}

int main()

{

#ifndef ONLINE_JUDGE

    freopen("bzoj4311.in","r",stdin);

    freopen("bzoj4311.out","w",stdout);

    const char LL[]="%I64d\n";

#else

    const char LL[]="%lld\n";

#endif

    n=read();

    build(,,n);

    for (int i=;i<=n;i++)

    {

        int op=read();

        if (op==) t++,a[t].x=read(),a[t].y=read(),a[t].s=i;

        else if (op==) m++,q[m].x=read(),q[m].y=read(),q[m].s=i;

        else

        {

            int x=read();

            ins(,a[x].s,i,a[x]);

            a[x].x=a[x].y=;

        }

    }

    for (int i=;i<=t;i++)

    if (a[i].x) ins(,a[i].s,n,a[i]);

    make();

    for (int i=;i<=m;i++) printf(LL,query(,q[i].s,q[i]));

    return ;

}