bzoj4066 简单题（K-D tree）

强制在线，内存还很小，只能K-D Tree。空间O(n),时间复杂度 O(m∗m−−√) ？每累积了很多点以后就暴力重构一下，保证一下效率。

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
#define ll long long
#define inf 0x3f3f3f3f
#define N 200010
inline int read(){
    int x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9') x=x*10+ch-'0',ch=getchar();
    return x*f;
}
int n,D,rt=0,x1,x2,y1,y2;
ll ans=0;
struct point{
    int d[2],w;
    int& operator[](int x){return d[x];}
    friend bool operator<(point a,point b){return a[D]<b[D];}
}P[N];
struct node{
    point x;int lc,rc,mn[2],mx[2];ll sum;
}tr[N];
inline void update(int p){
    int l=tr[p].lc,r=tr[p].rc;
    for(int i=0;i<2;++i){
        if(l){
            tr[p].mn[i]=min(tr[p].mn[i],tr[l].mn[i]);
            tr[p].mx[i]=max(tr[p].mx[i],tr[l].mx[i]);
        }if(r){
            tr[p].mn[i]=min(tr[p].mn[i],tr[r].mn[i]);
            tr[p].mx[i]=max(tr[p].mx[i],tr[r].mx[i]);
        }
    }tr[p].sum=tr[p].x.w+tr[l].sum+tr[r].sum;
}
inline void build(int &p,int l,int r,int op){
    int mid=l+r>>1;p=mid;tr[p].lc=tr[p].rc=0;D=op;
    nth_element(P+l,P+mid,P+r+1);tr[p].x=P[mid];tr[p].sum=P[mid].w;
    for(int i=0;i<2;++i) tr[p].mn[i]=tr[p].mx[i]=tr[p].x[i];
    if(l<mid) build(tr[p].lc,l,mid-1,op^1);
    if(r>mid) build(tr[p].rc,mid+1,r,op^1);
    update(p);
}
inline void ins(int &p,int k,int op){
    if(!p){
        p=k;tr[p].x=P[k];tr[p].sum=P[k].w;
        for(int i=0;i<2;++i) tr[p].mn[i]=tr[p].mx[i]=tr[p].x[i];return;
    }if(P[k][op]<=tr[p].x[op]) ins(tr[p].lc,k,op^1);
    else ins(tr[p].rc,k,op^1);update(p);
}
inline bool in(int p){
    return tr[p].mn[0]>=x1&&tr[p].mx[0]<=x2&&tr[p].mn[1]>=y1&&tr[p].mx[1]<=y2;
}
inline bool out(int p){
    return tr[p].mx[0]<x1||tr[p].mn[0]>x2||tr[p].mx[1]<y1||tr[p].mn[1]>y2;
}
inline bool jud(point x){
    return x1<=x[0]&&x[0]<=x2&&y1<=x[1]&&x[1]<=y2;
}
inline ll ask(int p){
    if(!p) return 0;
    if(in(p)) return tr[p].sum;
    if(out(p)) return 0;ll res=0;
    if(jud(tr[p].x)) res+=tr[p].x.w;
    return res+ask(tr[p].lc)+ask(tr[p].rc);
}
int main(){
// freopen("a.in","r",stdin);
    n=read();n=0;
    while(1){
        int op=read();if(op==3) break;
        if(op==1){
            P[++n][0]=read()^ans;P[n][1]=read()^ans;P[n].w=read()^ans;
            if(n%10000==0) rt=0,build(rt,1,n,0);
            else ins(rt,n,0);
        }else{
            x1=read()^ans;y1=read()^ans;x2=read()^ans;y2=read()^ans;
            printf("%lld\n",ans=ask(rt));
        }
    }return 0;
}