传送门

首先可以把约束条件看成一条直线，然后每个巧克力看成一个点，求给定区域内的点权和
用KDTree，每次判断一下当前矩形是否整个都在里面或都在外面，是的话直接返回，否则的话递归
注意，必须该矩形四个顶点都在里面或外面才能判断

//minamoto
#include<bits/stdc++.h>
#define R register
#define ll long long
#define fp(i,a,b) for(R int i=a,I=b+1;i<I;++i)
#define fd(i,a,b) for(R int i=a,I=b-1;i>I;--i)
#define go(u) for(int i=head[u],v=e[i].v;i;i=e[i].nx,v=e[i].v)
template<class T>inline bool cmin(T&a,const T&b){return a>b?a=b,1:0;}
template<class T>inline bool cmax(T&a,const T&b){return a<b?a=b,1:0;}
using namespace std;
char buf[1<<21],*p1=buf,*p2=buf;
inline char getc(){return p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<21,stdin),p1==p2)?EOF:*p1++;}
ll read(){R ll res,f=1;R char ch;while((ch=getc())>'9'||ch<'0')(ch=='-')&&(f=-1);for(res=ch-'0';(ch=getc())>='0'&&ch<='9';res=res*10+ch-'0');return res*f;
}
char sr[1<<21],z[20];int C=-1,Z=0;
inline void Ot(){fwrite(sr,1,C+1,stdout),C=-1;}
void print(R ll x){if(C>1<<20)Ot();if(x<0)sr[++C]='-',x=-x;while(z[++Z]=x%10+48,x/=10);while(sr[++C]=z[Z],--Z);sr[++C]='\n';
}
const int N=50005;
struct node{int v[2],mn[2],mx[2],w,l,r;ll sum;}tr[N];
int n,m,K,rt,a,b;ll c;
inline bool operator <(const node &a,const node &b){return a.v[K]<b.v[K];}
void upd(int p){int l=tr[p].l,r=tr[p].r;fp(i,0,1){tr[p].mn[i]=tr[p].mx[i]=tr[p].v[i];if(l)cmin(tr[p].mn[i],tr[l].mn[i]),cmax(tr[p].mx[i],tr[l].mx[i]);if(r)cmin(tr[p].mn[i],tr[r].mn[i]),cmax(tr[p].mx[i],tr[r].mx[i]);}tr[p].sum=tr[l].sum+tr[r].sum+tr[p].w;
}
int build(int l,int r,int k){K=k;int mid=(l+r)>>1;nth_element(tr+l,tr+mid,tr+r+1);if(l<mid)tr[mid].l=build(l,mid-1,k^1);if(mid<r)tr[mid].r=build(mid+1,r,k^1);upd(mid);return mid;
}
inline bool in(int x,int y){return 1ll*a*x+1ll*b*y<c;}
ll query(int p){if(!p)return 0;if(in(tr[p].mx[0],tr[p].mx[1])&&in(tr[p].mn[0],tr[p].mn[1])&&in(tr[p].mn[0],tr[p].mx[1])&&in(tr[p].mx[0],tr[p].mn[1]))return tr[p].sum;ll res=0;if(!in(tr[p].mn[0],tr[p].mn[1])&&!in(tr[p].mx[0],tr[p].mx[1])&&!in(tr[p].mn[0],tr[p].mx[1])&&!in(tr[p].mx[0],tr[p].mn[1]))return 0;if(in(tr[p].v[0],tr[p].v[1]))res+=tr[p].w;res+=query(tr[p].l)+query(tr[p].r);return res;
}
int main(){
//  freopen("testdata.in","r",stdin);n=read(),m=read();fp(i,1,n)tr[i].v[0]=read(),tr[i].v[1]=read(),tr[i].w=read();rt=build(1,n,0);while(m--)a=read(),b=read(),c=read(),print(query(rt));return Ot(),0;
}