P4148 简单题（KDTree）

传送门

KDTree
修改权值当做插入节点，不平衡就暴力重构，询问的时候判断当前节点代表的矩形是否在询问的矩形的，是的话返回答案，相离返回0，否则的话判断当前点是否在矩形内，然后继续递归下去

//minamoto
#include<bits/stdc++.h>
#define R register
#define fp(i,a,b) for(R int i=a,I=b+1;i<I;++i)
#define fd(i,a,b) for(R int i=a,I=b-1;i>I;--i)
#define go(u) for(int i=head[u],v=e[i].v;i;i=e[i].nx,v=e[i].v)
template<class T>inline bool cmax(T&a,const T&b){return a<b?a=b,1:0;}
template<class T>inline bool cmin(T&a,const T&b){return a>b?a=b,1:0;}
using namespace std;
char buf[1<<21],*p1=buf,*p2=buf;
inline char getc(){return p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<21,stdin),p1==p2)?EOF:*p1++;}
int read(){
    R int res,f=1;R char ch;
    while((ch=getc())>'9'||ch<'0')(ch=='-')&&(f=-1);
    for(res=ch-'0';(ch=getc())>='0'&&ch<='9';res=res*10+ch-'0');
    return res*f;
}
char sr[1<<21],z[20];int C=-1,Z=0;
inline void Ot(){fwrite(sr,1,C+1,stdout),C=-1;}
void print(R int x){
    if(C>1<<20)Ot();if(x<0)sr[++C]='-',x=-x;
    while(z[++Z]=x%10+48,x/=10);
    while(sr[++C]=z[Z],--Z);sr[++C]='\n';
}
const int N=2e5+5;
struct point{int v[2],w;}pool[N];
struct node{int mn[2],mx[2],sum,l,r,sz;point tp;}tr[N];
int n,m,K,rt,top,cnt,st[N],ans;
inline bool operator <(const point &a,const point &b){return a.v[K]<b.v[K];}
inline int newnode(){return top?st[top--]:++cnt;}
void upd(int p){
    int l=tr[p].l,r=tr[p].r;
    fp(i,0,1){
        tr[p].mn[i]=tr[p].mx[i]=tr[p].tp.v[i];
        if(l)cmin(tr[p].mn[i],tr[l].mn[i]),cmax(tr[p].mx[i],tr[l].mx[i]);
        if(r)cmin(tr[p].mn[i],tr[r].mn[i]),cmax(tr[p].mx[i],tr[r].mx[i]);
    }tr[p].sum=tr[l].sum+tr[r].sum+tr[p].tp.w,tr[p].sz=tr[l].sz+tr[r].sz+1;
}
int build(int l,int r,int k){
    if(l>r)return 0;int mid=(l+r)>>1,p=newnode();
    K=k,nth_element(pool+l,pool+mid,pool+r+1),tr[p].tp=pool[mid];
    tr[p].l=build(l,mid-1,k^1),tr[p].r=build(mid+1,r,k^1);
    upd(p);return p;
}
void pia(int p,int sum){
    if(tr[p].l)pia(tr[p].l,sum);
    pool[sum+tr[tr[p].l].sz+1]=tr[p].tp,st[++top]=p;
    if(tr[p].r)pia(tr[p].r,sum+tr[tr[p].l].sz+1);
}
void ck(int &p,int k){
    if(tr[p].sz*0.75<tr[tr[p].l].sz||tr[p].sz*0.75<tr[tr[p].r].sz)
    pia(p,0),p=build(1,tr[p].sz,k);
}
void ins(int &p,point res,int k){
    if(!p)return (void)(p=newnode(),tr[p].l=tr[p].r=0,tr[p].tp=res,upd(p));
    res.v[k]<=tr[p].tp.v[k]?ins(tr[p].l,res,k^1):ins(tr[p].r,res,k^1);
    upd(p),ck(p,k);
}
inline bool in(int x1,int y1,int x2,int y2,int X1,int Y1,int X2,int Y2){
    return x1<=X1&&y1<=Y1&&x2>=X2&&y2>=Y2;
}
inline bool out(int x1,int y1,int x2,int y2,int X1,int Y1,int X2,int Y2){
    return x1>X2||x2<X1||y1>Y2||y2<Y1;
}
int query(int p,int x1,int y1,int x2,int y2){
    if(!p)return 0;int res=0;
    if(in(x1,y1,x2,y2,tr[p].mn[0],tr[p].mn[1],tr[p].mx[0],tr[p].mx[1]))return tr[p].sum;
    if(out(x1,y1,x2,y2,tr[p].mn[0],tr[p].mn[1],tr[p].mx[0],tr[p].mx[1]))return 0;
    if(in(x1,y1,x2,y2,tr[p].tp.v[0],tr[p].tp.v[1],tr[p].tp.v[0],tr[p].tp.v[1]))res+=tr[p].tp.w;
    res+=query(tr[p].l,x1,y1,x2,y2)+query(tr[p].r,x1,y1,x2,y2);return res;
}
int main(){
//  freopen("testdata.in","r",stdin);
    n=read();int op,x1,y1,x2,y2;
    while(true){
        op=read();if(op==3)break;
        if(op==1)ins(rt,(point){read()^ans,read()^ans,read()^ans},0);
        else{
            x1=read()^ans,y1=read()^ans,x2=read()^ans,y2=read()^ans;
            ans=query(rt,x1,y1,x2,y2),print(ans);
        }
    }return Ot(),0;
}

原文地址：https://www.cnblogs.com/bztMinamoto/p/10090792.html