Subject to the effect, and the seeking of the data in a given claim.
Very simple, as data for achievements, maintain a \ (sum \) delicious degrees and upper and lower bounds becomes sweeter, \ (Query \) judgment is not sweetness to meet customer requirements, if four kinds with \ (( mi [0] -> mx [
1], mi [0] -> mi [1], mx [0] -> mi [1], mx [0] -> mx [1]) \) meet the requirements, then the whole range of answers are all contributions, direct plus \ (sum \) can be.
Otherwise, the judge went to the point of the current wave can contribute, then left and right query.
#include<cstdio>
#include<iostream>
#include<cstring>
#include<algorithm>
using namespace std;
#define int long long
const int MAXN=5e5+10;
//挺简单的基础K-D_TREE
struct pt{
int x[2],cnt;
}p[MAXN];
struct node{
int siz,mi[2],mx[2],sum;
//sum维护区间和
pt c;
}tr[MAXN];
int n,m,rt,tot,D;
int ls[MAXN],rs[MAXN];
int operator<(pt a,pt b){return a.x[D]<b.x[D];}
inline void pushup(int x){
int l=ls[x],r=rs[x];
tr[x].siz=tr[l].siz+tr[r].siz+1;
tr[x].sum=tr[l].sum+tr[r].sum+tr[x].c.cnt;
for(int i=0;i<=1;++i){
tr[x].mi[i]=tr[x].mx[i]=tr[x].c.x[i];
if(l)tr[x].mi[i]=min(tr[x].mi[i],tr[l].mi[i]),tr[x].mx[i]=max(tr[x].mx[i],tr[l].mx[i]);
if(r)tr[x].mi[i]=min(tr[x].mi[i],tr[r].mi[i]),tr[x].mx[i]=max(tr[x].mx[i],tr[r].mx[i]);
}
//更新子树信息
}
int build(int l,int r,int d){
if(l>r)return 0;
int x=++tot,mid=l+r>>1;
D=d;nth_element(p+l,p+mid,p+r+1);
tr[x].c=p[mid];ls[x]=build(l,mid-1,d^1);
rs[x]=build(mid+1,r,d^1);pushup(x);return x;
}
int A,B,C;
inline bool check(int x,int y){return A*x+B*y<C;}
int query(int x){
//区间查询
int t=0,res=0;
t+=check(tr[x].mi[0],tr[x].mi[1]);
t+=check(tr[x].mi[0],tr[x].mx[1]);
t+=check(tr[x].mx[0],tr[x].mi[1]);
t+=check(tr[x].mx[0],tr[x].mx[1]);
if(t==4)return tr[x].sum;//如果这个区间完全在查询范围
else if(t==0)return 0;//完全不在
if(check(tr[x].c.x[0],tr[x].c.x[1]))res+=tr[x].c.cnt;//判断这个点在不在
if(ls[x])res+=query(ls[x]);//分成两边累加答案
if(rs[x])res+=query(rs[x]);
return res;
}
signed main(){
scanf("%lld%lld",&n,&m);
for(int i=1;i<=n;++i)
scanf("%lld%lld%lld",&p[i].x[0],&p[i].x[1],&p[i].cnt);
rt=build(1,n,0);
for(int i=1;i<=m;++i){
scanf("%lld%lld%lld",&A,&B,&C);
printf("%lld\n",query(rt));
}
return 0;
}