[bzoj4066]简单题

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要求支持两种操作1)给点(x,y)加上一个权值v 2）查询矩阵(x1,y1,x2,y2)的权值和。操作数最多200000，强制在线。

大力kdtree

#include<iostream>
#include<cstdio>
#include<algorithm> 
#define MN 200000
using namespace std;
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=0,last,F,m;
struct P{
    int d[2],mn[2],mx[2],l,r,sum,num;
    P(){}
    P(int x1,int y1,int x2,int y2){mn[0]=x1;mx[0]=x2;mn[1]=y1;mx[1]=y2;}
    P(int x,int y){d[0]=x;d[1]=y;}
    int&operator[](int x){return d[x];}
    bool operator <(const P&b)const{return d[F]<b.d[F];}
    bool operator == (const P&b)const{return d[0]==b.d[0]&&d[1]==b.d[1];}
    inline bool in(const P&b)const{return b.mn[0]>=mn[0]&&b.mx[0]<=mx[0]&&b.mn[1]>=mn[1]&&b.mx[1]<=mx[1];}
    inline bool out(const P&b)const{return mn[0]>b.mx[0]||mx[0]<b.mn[0]||mn[1]>b.mx[1]||mx[1]<b.mn[1];}
    inline bool in(const int&x,const int&y)const{return x>=mn[0]&&x<=mx[0]&&y>=mn[1]&&y<=mx[1];}
//    void print(){cout<<mn[0]<<" "<<mx[0]<<" "<<mn[1]<<" "<<mx[1]<<endl;}
}t[MN+5];

struct tree{
    P p[MN+5],T;
    void update(int x)
    {
        int l=p[x].l,r=p[x].r;
        for(int i=0;i<2;i++)
        {
            if(l) p[x].mn[i]=min(p[x].mn[i],p[l].mn[i]),
                  p[x].mx[i]=max(p[x].mx[i],p[l].mx[i]);
            if(r) p[x].mn[i]=min(p[x].mn[i],p[r].mn[i]),
                  p[x].mx[i]=max(p[x].mx[i],p[r].mx[i]);    
        }
        p[x].sum=p[l].sum+p[r].sum+p[x].num;
    }
    int build(int lt,int rt,bool now)
    {
        if(lt>rt)return 0;
        int mid=lt+rt>>1;
        F=now;nth_element(t+lt,t+mid,t+rt+1);
        p[mid]=t[mid];
        p[mid].mn[0]=p[mid].mx[0]=p[mid][0];
        p[mid].mn[1]=p[mid].mx[1]=p[mid][1];
        p[mid].l=build(lt,mid-1,now^1);
        p[mid].r=build(mid+1,rt,now^1);
        update(mid);
        return mid;
    }
    int query(int x)
    {
        if(!x)return 0;
        if(T.in(p[x])) return p[x].sum;
        if(T.out(p[x])) return 0;
        int sum=0;
        if(T.in(p[x].d[0],p[x].d[1])) sum+=p[x].num;
        sum+=query(p[x].l)+query(p[x].r);
        return sum;
    }
    void ins(int&x,int v,bool now)
    {
        if(!x){
            x=++n;p[x]=T;
            for(register int i=0;i<2;i++)
                p[x].mn[i]=p[x].mx[i]=p[x][i];
            p[x].sum=p[x].num=v;p[x].l=p[x].r=0;return;
        }
        if(p[x]==T){ p[x].sum+=v;p[x].num+=v;return;}
        if(T[now]<p[x][now]) ins(p[x].l,v,now^1);
        else ins(p[x].r,v,now^1);
        update(x); 
    }
}tree;
int rt;
int main()
{
    read();
    for(register int i=1;;)
    {
        int opt=read();if(opt>2)break;
        if(opt==1)
        {
            int x=read()^last,y=read()^last,v=read()^last;tree.T=P(x,y);
            tree.ins(rt,v,0);
            if(++i==10000)
            {
                for(int j=1;j<=n;j++) t[j]=tree.p[j];
                rt=tree.build(1,n,0);i=0;
            }
        }
        else
        {
            int x1=read()^last,y1=read()^last,x2=read()^last,y2=read()^last;
            tree.T=P(x1,y1,x2,y2);
            printf("%d\n",last=tree.query(rt));
        }
    }
    return 0;
}