bzoj4066: 简单题

链接

bzoj

思路

强制在线。k-dtree。
卡常啊。空间开1e6就T了。

代码

#include <bits/stdc++.h>
#define my_min(a,b) (a<b?a:a=b)
#define my_max(a,b) (a>b?a:a=b)
using namespace std;
const double alpha=0.75;
const int N=2e5+7;
int read() {
    int x=0,f=1;
    char s=getchar();
    for(; s>'9'||s<'0'; s=getchar()) if(s=='-') f=-1;
    for(; s>='0'&&s<='9'; s=getchar()) x=x*10+s-'0';
    return x*f;
}
int WD,lastans;
struct point {
    int x[2],v;
    bool operator < (const point &b) const {
        return x[WD]<b.x[WD];
    }
} p[N];
struct node {
    int ls,rs,mi[2],ma[2],sum,siz;
    point a;
} e[N];
int rub[N],top,cnt;
int newnode() {
    if(top) return rub[top--];
    return ++cnt;
}
void update(int x,int y) {
    my_min(e[x].mi[0],e[y].mi[0]);
    my_min(e[x].mi[1],e[y].mi[1]);
    my_max(e[x].ma[0],e[y].ma[0]);
    my_max(e[x].ma[1],e[y].ma[1]);
}
void up(int u) {
    int ls=e[u].ls,rs=e[u].rs;
    e[u].mi[0]=e[u].ma[0]=e[u].a.x[0];
    e[u].mi[1]=e[u].ma[1]=e[u].a.x[1];
    e[u].sum=e[ls].sum+e[rs].sum+e[u].a.v;
    e[u].siz=e[ls].siz+e[rs].siz+1;
    if(ls) update(u,ls);
    if(rs) update(u,rs);
}
int build(int l,int r,int wd) {
    if(l>r) return 0;
    int mid=(l+r)>>1;
    WD=wd;
    nth_element(p+l,p+mid,p+r+1);
    int u=newnode();
    e[u].a=p[mid];
    e[u].ls=build(l,mid-1,wd^1);
    e[u].rs=build(mid+1,r,wd^1);
    up(u);
    return u;
}
void pia(int u,int num) {
    if(e[u].ls) pia(e[u].ls,num);
    p[num+e[e[u].ls].siz+1]=e[u].a;
    rub[++top]=u;
    if(e[u].rs) pia(e[u].rs,num+e[e[u].ls].siz+1);
}
int check(int &u,int wd) {
    if(alpha*e[u].siz<e[e[u].ls].siz||alpha*e[u].siz<e[e[u].rs].siz)
        pia(u,0),u=build(1,e[u].siz,wd);
}
void insert(point a,int &rt,int wd) {
    if(!rt) {
        rt=newnode();
        e[rt].a=a;
        e[rt].ls=e[rt].rs=0;
        up(rt);
        return;
    }
    if(e[rt].a.x[wd]<a.x[wd]) insert(a,e[rt].rs,wd^1);
    else insert(a,e[rt].ls,wd^1);
    up(rt);
    check(rt,wd);
}
int pd(node a,int x,int y,int X,int Y) {
    if(x<=a.mi[0]&&a.ma[0]<=X&&y<=a.mi[1]&&a.ma[1]<=Y) return 1;
    if(a.ma[0]<x||X<a.mi[0]) return 0;
    if(a.ma[1]<y||Y<a.mi[1]) return 0;
    return 2;
}
void query(int rt,int x,int y,int X,int Y) {
    if(x<=e[rt].a.x[0]&&e[rt].a.x[0]<=X
            &&y<=e[rt].a.x[1]&&e[rt].a.x[1]<=Y) lastans+=e[rt].a.v;
    int ls=e[rt].ls,rs=e[rt].rs;
    if(ls) {
        int PD=pd(e[ls],x,y,X,Y);
        if(PD==1) lastans+=e[ls].sum;
        if(PD==2) query(ls,x,y,X,Y);
    }
    if(rs) {
        int PD=pd(e[rs],x,y,X,Y);
        if(PD==1) lastans+=e[rs].sum;
        if(PD==2) query(rs,x,y,X,Y);
    }
}
int main() {
    int rt=0,n=read();
    while(233) {
        int opt=read();
        if(opt==1) {
            point a;
            a.x[0]=read()^lastans,a.x[1]=read()^lastans,a.v=read()^lastans;
            insert(a,rt,0);
        } else if(opt==2) {
            int x=read()^lastans,y=read()^lastans,X=read()^lastans,Y=read()^lastans;
            lastans=0;
            query(rt,x,y,X,Y);
            printf("%d\n",lastans);
        } else break;
    }
    return 0;
}