BZOJ 4066 简单题（KD树）

#include <cstdio>
#include <algorithm>
using namespace std;
const int N=300000,INF=1e9;
namespace KD_Tree{
    struct Dot{  
        int d[2],mn[2],mx[2],l,r,sz,sum,val;  
        Dot(){l=r=0;}  
        Dot(int x,int y,int c){d[0]=x;d[1]=y;val=c;l=r=sz=sum=0;}  
        int& operator [] (int x){return d[x];}  
    };  
    int D,dcnt=0,pt[N]; 
    Dot T[N];
    inline void umax(int&a,int b){if(a<b)a=b;}
    inline void umin(int&a,int b){if(a>b)a=b;}
    inline bool cmp(int x,int y){return T[x][D]<T[y][D];}
    inline void up(int x){
        T[x].sz=T[T[x].l].sz+T[T[x].r].sz+1;
        T[x].sum=T[T[x].l].sum+T[T[x].r].sum+T[x].val;
        T[x].mn[0]=T[x].mx[0]=T[x][0];  
        T[x].mn[1]=T[x].mx[1]=T[x][1]; 
        if(T[x].l){
            umax(T[x].mx[0],T[T[x].l].mx[0]);
            umin(T[x].mn[0],T[T[x].l].mn[0]);   
            umax(T[x].mx[1],T[T[x].l].mx[1]);
            umin(T[x].mn[1],T[T[x].l].mn[1]);
        }
        if(T[x].r){
            umax(T[x].mx[0],T[T[x].r].mx[0]);
            umin(T[x].mn[0],T[T[x].r].mn[0]);
            umax(T[x].mx[1],T[T[x].r].mx[1]);
            umin(T[x].mn[1],T[T[x].r].mn[1]);
        }
    }
    inline int NewDot(int x,int y,int c){
        ++dcnt; pt[dcnt]=dcnt;
        T[dcnt][0]=x; T[dcnt][1]=y; T[dcnt].val=c;
        return up(dcnt),dcnt;
    }
    // 查询矩阵内数字和
    int query(int x,int x0,int y0,int x1,int y1){
        if(!x||T[x].mn[0]>x1||T[x].mx[0]<x0||T[x].mn[1]>y1||T[x].mx[1]<y0)return 0;  
        if(T[x].mn[0]>=x0&&T[x].mx[0]<=x1&&T[x].mn[1]>=y0&&T[x].mx[1]<=y1)return T[x].sum;  
        int res=0;  
        if(T[x][0]>=x0&&T[x][0]<=x1&&T[x][1]>=y0&&T[x][1]<=y1)res+=T[x].val;  
        return res+query(T[x].l,x0,y0,x1,y1)+query(T[x].r,x0,y0,x1,y1); 
    }
    void Insert(int&x,int D,const Dot&p){  
        if(!x){x=NewDot(p.d[0],p.d[1],p.val);return;}  
        if(p.d[D]<T[x][D])Insert(T[x].l,D^1,p);  
        else Insert(T[x].r,D^1,p);  
        up(x);  
    }
    // 暴力重构
    int Rebuild(int l,int r,int now){  
        if(l>r)return 0;  
        int mid=(l+r)>>1,x;  
        D=now;  
        nth_element(pt+l,pt+mid,pt+r+1,cmp);  
        x=pt[mid];  
        T[x].l=Rebuild(l,mid-1,now^1);  
        T[x].r=Rebuild(mid+1,r,now^1);  
        return up(x),x;  
    } 
}
int n,op,x0,y0,x1,y1,c,ans=0,root=0;
int main(){
    scanf("%d",&n);
    while(scanf("%d",&op)){
        if(op==3)break;
        if(op==1){
            scanf("%d%d%d",&x0,&y0,&c);
            x0^=ans,y0^=ans,c^=ans;
            KD_Tree::Insert(root,0,KD_Tree::Dot(x0,y0,c));
            if(KD_Tree::dcnt%5000==0)root=KD_Tree::Rebuild(1,KD_Tree::dcnt,0);
        }else{
            scanf("%d%d%d%d",&x0,&y0,&x1,&y1);
            x0^=ans,y0^=ans,x1^=ans,y1^=ans;
            printf("%d\n",ans=KD_Tree::query(root,x0,y0,x1,y1));
        }
    }return 0;
}