[bzoj 4066]简单题

传送门

Description

两个操作，往一个格子里加一个数和求给定矩形的权值和，强制在线，操作数\(\leq 200000\)

Solution 

直接上KD-tree

为了保证树的形态较为优美

每加入\(10000\)个数后，对KD-tree进行重构

Code 

#include<bits/stdc++.h>
#define ll long long
#define max(a,b) ((a)>(b)?(a):(b))
#define min(a,b) ((a)<(b)?(a):(b))
#define reg register
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<<3)+(x<<1)+ch-'0';ch=getchar();}
	return x*f;
}
const int MN=2e5+5;
int N,F;
struct Point{
	int d[2],mn[2],mx[2],l,r,sum,val;
	Point(){}
	Point(int x,int y){d[0]=x,d[1]=y;}
	Point(int x,int y,int x_,int y_){mn[0]=x,mn[1]=y;mx[0]=x_,mx[1]=y_;}
	int&operator[](int x){return d[x];}
	bool operator<(const Point&b)const{return d[F]<b.d[F];}
    bool operator==(const Point&b)const{return d[0]==b.d[0]&&d[1]==b.d[1];}
    inline bool in(const Point&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 Point&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];}
}a[MN];
struct KD_tree{
    Point p[MN],T;
    void up(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].val;
    }
    int build(int l,int r,int th)
    {
        if(l>r)return 0;int mid=l+r>>1;
        F=th;std::nth_element(a+l,a+mid,a+r+1);
        p[mid]=a[mid];
        p[mid].mn[0]=p[mid].mx[0]=a[mid][0];
        p[mid].mn[1]=p[mid].mx[1]=a[mid][1];
        p[mid].l=build(l,mid-1,th^1);
		p[mid].r=build(mid+1,r,th^1);
        up(mid);
        return mid;
    }
    int qry(int x)
    {
        if(!x)return 0;
        if(T.in(p[x])) return p[x].sum;
        if(T.out(p[x])) return 0;
        int ret=0;
        if(T.in(p[x].d[0],p[x].d[1])) ret+=p[x].val;
        ret+=qry(p[x].l)+qry(p[x].r);
        return ret;
    }
    void ins(int&x,int v,int th)
    {
        if(!x)
		{
            x=++N;p[x]=T;
            for(reg int i=0;i<2;++i)
                p[x].mn[i]=p[x].mx[i]=p[x][i];
            p[x].sum=p[x].val=v;p[x].l=p[x].r=0;return;
        }
        if(p[x]==T){p[x].sum+=v;p[x].val+=v;return;}
        if(T[th]<p[x][th]) ins(p[x].l,v,th^1);
        else ins(p[x].r,v,th^1);
        up(x); 
    }
}kdtree;
int rt;
int main()
{
	read();
	reg int opt,x,y,v,x_,y_,la=0;
	for(reg int i=1,j;;)
	{
		opt=read();if(opt==3)break;
		if(opt==1)
		{
			x=read()^la,y=read()^la,v=read()^la;
			kdtree.T=Point(x,y);
			kdtree.ins(rt,v,0);
			if(++i==10000)
			{
				for(j=1;j<=N;++j) a[j]=kdtree.p[j];
                rt=kdtree.build(1,N,0);i=0;
			}
		}
		if(opt==2)
		{
			x=read()^la,y=read()^la,x_=read()^la,y_=read()^la;
			kdtree.T=Point(x,y,x_,y_);
			printf("%d\n",la=kdtree.qry(rt));
		}
	}
	return 0;
}

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