可以發現答案一定在所有向量終點形成的上凸殼上，於是在上凸殼上三分即可。

對於刪除操作，相當於每個向量有一個作用區間，線段樹分治即可。$O(n\log^2 n)$

同時可以發現，當詢問按斜率排序後，每個凸殼上的決策點也是單調變化的，於是可以記錄每次的決策位置。$O(n\log n)$

$O(n\log^2 n)$:

 1 #include<cstdio>
 2 #include<vector>
 3 #include<cstring>
 4 #include<algorithm>
 5 #define ls (x<<1)
 6
 
 #define rs (ls|1)
 7 #define lson ls,L,mid
 8 #define rson rs,mid+1,R
 9 #define rep(i,l,r) for (int i=(l); i<=(r); i++)
10 typedef long long ll;
11 using namespace std;
12 
13 const int N=200010;
14 int n,op,x,y,tim,tot,d[N<<2];
15 struct P{ ll x,y; };
16 struct D{ int l,r; P p; }p[N],q[N];
 
17 vector<P>v[N<<2];
18 P operator -(P a,P b){ return (P){a.x-b.x,a.y-b.y}; }
19 ll operator *(P a,P b){ return a.x*b.y-a.y*b.x; }
20 bool cmp1(const D &a,const D &b){ return (a.p.x==b.p.x) ? a.p.y<b.p.y : a.p.x<b.p.x; }
21 bool cmp2(const D &a,const D &b){ return
 
 a.p*b.p<0; }
22 ll calc(P a,P b){ return a.x*b.x+a.y*b.y; }
23 
24 void ins(int x,int L,int R,int l,int r,P p){
25     if (l<=L && R<=r){
26         while (v[x].size()>1 && (v[x][v[x].size()-1]-v[x][v[x].size()-2])*(p-v[x][v[x].size()-2])>=0) v[x].pop_back();
27         v[x].push_back(p); return;
28     }
29     int mid=(L+R)>>1;
30     if (l<=mid) ins(lson,l,r,p);
31     if (r>mid) ins(rson,l,r,p);
32 }
33 
34 ll que(int x,int L,int R,int pos,P p){
35     ll ans=0;
36     if (v[x].size()){
37         int l=0,r=v[x].size()-1;
38         while (l+2<r){
39             int m1=l+(r-l)/3,m2=r-(r-l)/3;
40             if (calc(p,v[x][m1])>calc(p,v[x][m2])) r=m2; else l=m1;
41         }
42         rep(i,l,r) ans=max(ans,calc(p,v[x][i]));
43     }
44     if (L==R) return ans;
45     int mid=(L+R)>>1;
46     if (pos<=mid) return max(ans,que(lson,pos,p)); else return max(ans,que(rson,pos,p));
47 }
48 
49 int main(){
50     freopen("bzoj4311.in","r",stdin);
51     freopen("bzoj4311.out","w",stdout);
52     scanf("%d",&n);
53     rep(i,1,n){
54         scanf("%d",&op);
55         if (op==1) scanf("%d%d",&x,&y),p[++tim]=(D){i,n,x,y};
56         if (op==2) scanf("%d",&x),p[x].r=i;
57         if (op==3) scanf("%d%d",&x,&y),q[++tot]=(D){i,tot,x,y};
58     }
59     sort(p+1,p+tim+1,cmp1);
60     rep(i,1,tim) ins(1,1,n,p[i].l,p[i].r,p[i].p);
61     rep(i,1,tot) printf("%lld\n",que(1,1,n,q[i].l,q[i].p));
62     return 0;
63 }

$O(n\log n)$

 1 #include<cstdio>
 2 #include<vector>
 3 #include<cstring>
 4 #include<algorithm>
 5 #define ls (x<<1)
 6 #define rs (ls|1)
 7 #define lson ls,L,mid
 8 #define rson rs,mid+1,R
 9 #define rep(i,l,r) for (int i=(l); i<=(r); i++)
10 typedef long long ll;
11 using namespace std;
12 
13 const int N=200010;
14 ll ans[N];
15 int n,op,x,y,tim,tot,d[N<<2];
16 struct P{ ll x,y; };
17 struct D{ int l,r; P p; }p[N],q[N];
18 vector<P>v[N<<2];
19 P operator -(P a,P b){ return (P){a.x-b.x,a.y-b.y}; }
20 ll operator *(P a,P b){ return a.x*b.y-a.y*b.x; }
21 bool cmp1(const D &a,const D &b){ return (a.p.x==b.p.x) ? a.p.y<b.p.y : a.p.x<b.p.x; }
22 bool cmp2(const D &a,const D &b){ return a.p*b.p<0; }
23 ll calc(P a,P b){ return a.x*b.x+a.y*b.y; }
24 
25 void ins(int x,int L,int R,int l,int r,P p){
26     if (l<=L && R<=r){
27         while (v[x].size()>1 && (v[x][v[x].size()-1]-v[x][v[x].size()-2])*(p-v[x][v[x].size()-2])>=0) v[x].pop_back();
28         v[x].push_back(p); return;
29     }
30     int mid=(L+R)>>1;
31     if (l<=mid) ins(lson,l,r,p);
32     if (r>mid) ins(rson,l,r,p);
33 }
34 
35 ll que(int x,int L,int R,int pos,P p){
36     ll ans=0;
37     if (v[x].size()){
38         while (d[x]<(int)v[x].size()-1 && calc(p,v[x][d[x]+1])>=calc(p,v[x][d[x]])) d[x]++;
39         ans=calc(p,v[x][d[x]]);
40     }
41     if (L==R) return ans;
42     int mid=(L+R)>>1;
43     if (pos<=mid) return max(ans,que(lson,pos,p)); else return max(ans,que(rson,pos,p));
44 }
45 
46 int main(){
47     freopen("bzoj4311.in","r",stdin);
48     freopen("bzoj4311.out","w",stdout);
49     scanf("%d",&n);
50     rep(i,1,n){
51         scanf("%d",&op);
52         if (op==1) scanf("%d%d",&x,&y),p[++tim]=(D){i,n,x,y};
53         if (op==2) scanf("%d",&x),p[x].r=i;
54         if (op==3) scanf("%d%d",&x,&y),q[++tot]=(D){i,tot,x,y};
55     }
56     sort(p+1,p+tim+1,cmp1);
57     rep(i,1,tim) ins(1,1,n,p[i].l,p[i].r,p[i].p);
58     sort(q+1,q+tot+1,cmp2);
59     rep(i,1,tot) ans[q[i].r]=que(1,1,n,q[i].l,q[i].p);
60     rep(i,1,tot) printf("%lld\n",ans[i]);
61     return 0;
62 }