传送门

分析

我们多维护一个值,代表某个点子树中所有点的权值和

于是如果某个点它的min和max乘a(/b)的值小于范围则直接把整个子树都加进去

估价函数就是这个点的子树中的理论最小值

代码

#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<algorithm>
#include<cctype>
#include<cmath>
#include<cstdlib>
#include<ctime>
#include<queue>
#include<vector>
#include<set>
#include<map>
#include<stack>
using namespace std;
#define int long long
const int inf = 1e9;
inline int ra(){
int x=,f=;char s=getchar();
while(!isdigit(s)){if(s=='-')f=-;s=getchar();}
while(isdigit(s))x=(x<<)+(x<<)+(s-''),s=getchar();
return x*f;
}
struct kd {
int d[],mx[],mn[],le,ri,id,sum,val;
};
kd t[],now;
int n,m,root,wh,Ans;
inline bool operator < (kd a,kd b){
return a.d[wh]<b.d[wh];
}
inline void up(int rt){
for(int i=;i<;++i){
t[rt].mn[i]=min(t[rt].mn[i],min(t[t[rt].le].mn[i],t[t[rt].ri].mn[i]));
t[rt].mx[i]=max(t[rt].mx[i],max(t[t[rt].le].mx[i],t[t[rt].ri].mx[i]));
}
t[rt].sum=t[t[rt].le].sum+t[t[rt].ri].sum+t[rt].val;
}
inline void build(int &x,int le,int ri,int wwh){
wh=wwh;
int mid=(le+ri)>>;
x=mid;
nth_element(t+le,t+x,t+ri+);
for(int i=;i<;++i)
t[x].mn[i]=t[x].mx[i]=t[x].d[i];
if(le<x)build(t[x].le,le,mid-,wwh^);
if(ri>x)build(t[x].ri,mid+,ri,wwh^);
up(x);
}
inline int getd(kd a,kd b){
if(!a.id)return inf;
int res=;
for(int i=;i<;++i)res+=a.d[i]*b.d[i];
return res;
}
inline int calc(int x){
if(!x)return inf;
int res=;
for(int i=;i<;++i)
res+=min(t[x].mn[i]*now.d[i],t[x].mx[i]*now.d[i]);
return res;
}
inline int getmax(int x){
if(!x)return inf;
int res=;
for(int i=;i<;++i)
res+=max(t[x].mn[i]*now.d[i],t[x].mx[i]*now.d[i]);
return res;
}
inline void qurey(int x){
if(!x)return;
if(getmax(x)<now.val){
Ans+=t[x].sum;
return;
}
int dl=calc(t[x].le),dr=calc(t[x].ri),d=getd(t[x],now);
if(d<now.val)Ans+=t[x].val;
if(dl<now.val)qurey(t[x].le);
if(dr<now.val)qurey(t[x].ri);
}
signed main(){
int i,j,k;
t[].mn[]=t[].mn[]=inf;
t[].mx[]=t[].mx[]=-inf;
scanf("%lld%lld",&n,&m);
for(i=;i<=n;++i){
t[i].d[]=ra(),t[i].d[]=ra(),t[i].val=ra();
t[i].id=i;
}
build(root,,n,);
for(i=;i<=m;++i){
Ans=;
now.d[]=ra(),now.d[]=ra(),now.val=ra();
qurey(root);
printf("%lld\n",Ans);
}
return ;
}
05-11 22:13