1 #include <cmath>
2 #include <ctime>
3 #include <cstdio>
4 #include <cstdlib>
5 #include <cstring>
6 #include <iostream>
7 #include <algorithm>
8 # define maxn 30010
9 using namespace std;
10 void ot(int x){cout<<"***** ->"<<x<<endl;}
11 int n,m;
12 int bd[maxn];
13 int ans1[maxn],ans2[maxn]; //ans2==num
14 struct ROU{
15 int val,id;
16 };
17 bool operator < (const ROU a,const ROU b){
18 if(a.val==b.val) return a.id<b.id;
19 return a.val<b.val;
20 }
21 struct Treap{
22 Treap* ch[2];
23 ROU tt;
24 int size,key,mk1,mk2;
25 Treap(int v,int ii)
26 {size=1; tt.val=v; tt.id=ii; key=rand(); mk1=mk2=0; ch[0]=ch[1]=NULL;}
27 void update()
28 {size=1+(ch[0]?ch[0]->size:0)+(ch[1]?ch[1]->size:0);}
29 };
30 typedef pair<Treap*,Treap*> D;
31 int Size(Treap* now){return now? now->size:0;}
32 int mmxx(Treap *now){return now? now->tt.val:0;}
33 void pushdown(Treap* &now){
34 if(!now) return;
35 if(now->ch[0]){
36 now->ch[0]->mk1=max(now->mk1,now->ch[0]->mk1);
37 now->ch[0]->mk2=max(now->mk2,now->ch[0]->mk2);
38 }
39 if(now->ch[1]){
40 now->ch[1]->mk1=max(now->mk1,now->ch[1]->mk1);
41 now->ch[1]->mk2=max(now->mk2,now->ch[1]->mk2);
42 }
43 int ii=now->tt.id;
44 ans1[ii]=max(ans1[ii],now->mk1);
45 ans2[ii]=max(ans2[ii],now->mk2);
46 now->mk1=0; now->mk2=0;
47 }
48 D Split(Treap* now,int k){
49 if(!now) return D(NULL,NULL);
50 D y;
51 pushdown(now);
52 if(Size(now->ch[0])>=k)
53 {y=Split(now->ch[0],k); now->ch[0]=y.second; now->update(); y.second=now;}
54 else
55 {y=Split(now->ch[1],k-Size(now->ch[0])-1); now->ch[1]=y.first; now->update(); y.first=now;}
56 return y;
57 }
58 Treap* Merge(Treap* a,Treap* b){
59 if(!a) return b;
60 if(!b) return a;
61 if(a->key < b->key)
62 {a->ch[1]=Merge(a->ch[1],b); a->update(); return a;}
63 else
64 {b->ch[0]=Merge(a,b->ch[0]); b->update(); return b;}
65 }
66 int Getkth(Treap* now,ROU o){
67 if(!now) return 0;
68 return (!(now->tt<o))? Getkth(now->ch[0],o):Getkth(now->ch[1],o)+Size(now->ch[0])+1;
69 }
70 int Findkth(Treap* &rt,int k){
71 D x=Split(rt,k-1);
72 D y=Split(x.second,1);
73 Treap* ans=y.first;
74 rt=Merge(Merge(x.first,ans),y.second);
75 return ans? ans->tt.val:0;
76 }
77 void add_mk(ROU o,Treap* &now){
78 if(!now) return ;
79 int da=Findkth(now,now->size);
80 now->mk1=max(now->mk1,o.val);
81 now->mk2=max(now->mk2,now->size);
82 ans1[o.id]=max(ans1[o.id],da);
83 ans2[o.id]=max(ans2[o.id],now->size);
84 }
85 void Insert(ROU o,Treap* &rt){
86 add_mk(o,rt);
87 int k=Getkth(rt,o);
88 D x=Split(rt,k);
89 Treap* now=new Treap(o.val,o.id);
90 rt=Merge(Merge(x.first,now),x.second);
91 }
92 void Delete(ROU o,Treap* &rt){
93 int k=Getkth(rt,o);
94 D x=Split(rt,k);
95 D y=Split(x.second,1);
96 int ii=y.first->tt.id;
97 ans1[ii]=max(ans1[ii],y.first->mk1);
98 ans2[ii]=max(ans2[ii],y.first->mk2);
99 rt=Merge(x.first,y.second);
100 }
101 struct HHs{
102 struct node{
103 int x,y,nxt;
104 Treap* rt;
105 }g[4000000];
106 int e,adj[80000];
107 int modx;
108 int mody;
109 int mod;
110 HHs(){
111 memset(adj,-1,sizeof(adj));
112 modx=7307; mody=9991; mod=76543;
113 }
114 void add(int now,int x,int y){
115 g[e].x=x; g[e].y=y; g[e].nxt=adj[now];
116 adj[now]=e++;
117 }
118 int find(int x,int y){
119 int now=(((x%modx)+modx)%modx)*(((y%mody)+mody)%mody)%mod;
120 for(int i=adj[now];i!=-1;i=g[i].nxt){
121 if(x==g[i].x && y==g[i].y){
122 return i;
123 }
124 }
125 add(now,x,y); return e-1;
126 }
127 }hs;
128 int pos[maxn][3];
129 void init(){
130 scanf("%d",&n);
131 int now; ROU o;
132 for(int i=1;i<=n;i++){
133 scanf("%d%d%d",&bd[i],&pos[i][0],&pos[i][1]);
134 now=hs.find(pos[i][0],pos[i][1]);
135 o.id=i; o.val=bd[i];
136 Insert(o,hs.g[now].rt);
137 }
138 }
139 void work(){
140 scanf("%d",&m);
141 int iid,x,y;
142 int now; ROU o;
143 for(int i=1;i<=m;i++){
144 scanf("%d%d%d",&iid,&x,&y);
145 o.id=iid; o.val=bd[iid];
146 now=hs.find(pos[iid][0],pos[iid][1]);
147 Delete(o,hs.g[now].rt);
148 pos[iid][0]=x; pos[iid][1]=y;
149 now=hs.find(x,y);
150 Insert(o,hs.g[now].rt);
151 }
152 for(int i=1;i<=n;i++){
153 now=hs.find(pos[i][0],pos[i][1]);
154 o.id=i; o.val=bd[i];
155 Delete(o,hs.g[now].rt);
156 }
157 long long ans;
158 for(int i=1;i<=n;i++){
159 ans=(long long)ans1[i]*ans2[i];
160 printf("%lld\n",ans);
161 }
162 }
163 int main(){
164 // freopen("a.in","r",stdin);
165 init();
166 work();
167 }
