题解 P4848/LOJ6016/BZOJ4605【崂山白花蛇草水】

简化题面

维护一个二维平面，支持两种操作

\(1\)，插入一个节点坐标为\((x,y)\)且权值为\(v\)

\(2\)，查询矩形范围为\(x_0,x_1,y_0,y_1\)内第\(k\)大的节点权值为多少，若不够\(k\)个则输出NAIVE!ORZzyz.

强制在线。

数据范围\(n\le500000\)，\(q\le100000\)，\(1\le x, y\le n\)，\(1\le v\le 10^9\)，\(1\le x_1\le x_2\le n\)，\(1\le y_1\le y_2\le n\)，\(1\le k\le q\)。

解题思路

二维平面内插入节点查询范围内信息首先想到用\(\mathcal{KDtree}\)维护，但是考虑查询第\(k\)大如何进行维护，考虑树套树，里层为\(\mathcal{KDtree}\)外层为权值线段树，我们只用在权值线段树上对价值进行二分即可。

注意权值线段树需要动态开点要不然空间会炸掉，\(\mathcal{KDtree}\)维护节点插入的时候应当注意失衡时和替罪羊树一样重构即可。

复杂度我不太会分析，大概是一个插入复杂度\(O(nlog^2 nlogv)\)和一个查询复杂度\(O(n\sqrt nlogv)\)

\(\mathcal{Code}\)

// Author: Ame__
#include<bits/stdc++.h>
#include<stdint.h>
#define _ 0
#define AME__DEBUG
#define bomb exit(0)
#define LOG(FMT...) fprintf(stderr , FMT)
#define TOWA(FMT...) fprintf(stdout , FMT)
using namespace std;
/*Grievous Lady*/
    
typedef int32_t i32;
typedef int64_t i64;
typedef double qwq;
    
const int BUF_SIZE = 1 << 12;
char buf[BUF_SIZE] , *buf_s = buf , *buf_t = buf + 1;
    
#define PTR_NEXT() \
{ \
    buf_s ++; \
    if(buf_s == buf_t) \
    { \
        buf_s = buf; \
        buf_t = buf + fread(buf , 1 , BUF_SIZE , stdin); \
    } \
}
    
#define mians(_s_) \
{ \
    while(!isgraph(*buf_s)) PTR_NEXT();\
    char register *_ptr_ = (_s_); \
    while(isgraph(*buf_s) || *buf_s == '-') \
    { \
        *(_ptr_ ++) = *buf_s; \
        PTR_NEXT(); \
    } \
    (*_ptr_) = '\0'; \
}
    
template <typename _n_> void mian(_n_ & _x_){
    while(*buf_s != '-' && !isdigit(*buf_s)) PTR_NEXT();
    bool register _nega_ = false; if(*buf_s == '-'){ _nega_ = true; PTR_NEXT(); }
    _x_ = 0; while(isdigit(*buf_s)){ _x_ = _x_ * 10 + *buf_s - '0'; PTR_NEXT(); } if(_nega_) _x_ = -_x_;
}
    
const i32 INF = 0x3f3f3f3f;
const i32 maxv = 1e9;
const i32 kato = 5e5 + 10;

template <typename _n_> bool cmax(_n_ &a , const _n_ &b){ return a < b ? a = b , 1 : 0; }
template <typename _n_> bool cmin(_n_ &a , const _n_ &b){ return a > b ? a = b , 1 : 0; }
    
i32 n , q , x , y , z , t , k , opt , lastans = 0;

struct point{
    i32 x , y , val;
    point(i32 x = 0 , i32 y = 0 , i32 val = 0): x(x) , y(y) , val(val){ }
    friend bool operator !=(const point &a , const point &b){
        return a.x != b.x || a.y != b.y;
    }
}b[kato];

inline bool cmp1(const point &a , const point &b){
    return a.x < b.x;
}

inline bool cmp2(const point &a , const point &b){
    return a.y < b.y;
}

namespace K_D_tree{
    struct node{
        node *ls , *rs;
        static queue<node*> q;
        point p;
        i32 x1 , x2 , y1 , y2 , size;
        node(){ }
        node(const point &qaq): p(qaq){
            ls = rs = 0x0 , x1 = x2 = p.x , y1 = y2 = p.y , size = 1;
        }
        inline void up1(node *x){
            this -> x1 = min(this -> x1 , x -> x1) , this -> x2 = max(this -> x2 , x -> x2);
            this -> y1 = min(this -> y1 , x -> y1) , this -> y2 = max(this -> y2 , x -> y2);
        }
        inline void up2(){
            size = (this -> ls ? this -> ls -> size : 0) + (this -> rs ? this -> rs -> size : 0) + 1;
        }
        void *operator new(size_t){
            static node *S = 0x0 , *T = 0x0; node *tmp;
            return q.empty() ? (S == T && (T = (S = new node[1024]) + 1024 , S == T) ? 0x0 : S ++) : (tmp = q.front() , q.pop() , tmp); 
        }
        void operator delete(void *qaq){ q.push(static_cast<node*>(qaq)); }
    };

    queue<node*> node::q;

    i32 tot , top;

    inline void del(node *&o){
        if(o -> ls) del(o -> ls);
        if(o -> rs) del(o -> rs);
        b[++ tot] = o -> p;
    }

    inline node *build(node *fa , i32 l , i32 r , i32 opt , point *a){
        if(l > r) return 0x0;
        i32 mid = (l + r) >> 1;
        nth_element(a + l , a + mid , a + r + 1 , opt ? cmp2 : cmp1);
        node *o = new node(a[mid]);
        o -> ls = build(o , l , mid - 1 , opt ^ 1 , a);
        o -> rs = build(o , mid + 1 , r , opt ^ 1 , a);
        if(o -> ls) o -> up1(o -> ls);
        if(o -> rs) o -> up1(o -> rs);
        o -> up2();
        return o;
    }

    inline void judge(node *&o , i32 opt){
        tot = 0; i32 res = o -> size;
        if(0.725 * o -> size <= static_cast<qwq>(max(o -> ls ? o -> ls -> size : 0 , o -> rs ? o -> rs -> size : 0))) del(o) , o = build(0x0 , 1 , res , opt , b);
    }

    inline void insert(node *&o , const point &p , i32 opt){
        if(!o) return void(o = new node(p));
        if(opt == 1){
            if(p.y <= o -> p.y) insert(o -> ls , p , opt ^ 1);
            else insert(o -> rs , p , opt ^ 1);
        }else{
            if(p.x <= o -> p.x) insert(o -> ls , p , opt ^ 1);
            else insert(o -> rs , p , opt ^ 1);
        }
        if(o -> ls) o -> up1(o -> ls);
        if(o -> rs) o -> up1(o -> rs);
        o -> up2();
        judge(o , opt);
    }

    inline i32 ask(node *o , i32 x1 , i32 x2 , i32 y1 , i32 y2){
        if(!o || x2 < o -> x1 || x1 > o -> x2 || y2 < o -> y1 || y1 > o -> y2) return 0;
        if(x1 <= o -> x1 && o -> x2 <= x2 && y1 <= o -> y1 && o -> y2 <= y2) return o -> size;
        i32 res = 0;
        if(x1 <= o -> p.x && o -> p.x <= x2 && y1 <= o -> p.y && o -> p.y <= y2) res += 1;
        return res + ask(o -> ls , x1 , x2 , y1 , y2) + ask(o -> rs , x1 , x2 , y1 , y2);
    }
}

namespace Segement_Tree{
    struct node{
        node *ls , *rs;
        K_D_tree::node *p;
    }_pool[kato] = {{_pool , _pool , 0x0}} , *tail = _pool , *null = tail ++ , *root = null;
    
    inline void insert(node *&o , i32 l , i32 r , const point &p , i32 k){
        if(o == null) o = new(tail ++) node(*null);
        K_D_tree::insert(o -> p , p , 0); 
        if(l == r) return;
        i32 mid = (l + r) >> 1;
        if(k <= mid) insert(o -> ls , l , mid , p , k);
        else insert(o -> rs , mid + 1 , r , p , k);
    }

    inline i32 ask(node *o , i32 l , i32 r , i32 x1 , i32 x2 , i32 y1 , i32 y2 , i32 k){
        if(l == r) return l;
        i32 res = K_D_tree::ask(o -> rs -> p , x1 , x2 , y1 , y2);
        i32 mid = (l + r) >> 1;
        if(k > res) return ask(o -> ls , l , mid , x1 , x2 , y1 , y2 , k - res);
        else return ask(o -> rs , mid + 1 , r , x1 , x2 , y1 , y2 , k);
    }
}

inline int Ame_(){
#ifdef AME__
    freopen(".in" , "r" , stdin); freopen(".out" , "w" , stdout); int nol_cl = clock();
#endif
    mian(n) , mian(q);
    for(; q --> 0 ;){
        mian(opt);
        if(opt == 1){ mian(x) , mian(y) , mian(k) , x ^= lastans , y ^= lastans , k ^= lastans , Segement_Tree::insert(Segement_Tree::root , 0 , maxv , point(x , y) , k); }
        if(opt == 2){ mian(x) , mian(y) , mian(z) , mian(t) , mian(k) , x ^= lastans , y ^= lastans , z ^= lastans , t ^= lastans , k ^= lastans; if((lastans = Segement_Tree::ask(Segement_Tree::root , 0 , maxv , x , z , y , t , k))) TOWA("%d\n" , lastans); else TOWA("NAIVE!ORZzyz.\n"); }
    }
#ifdef AME__TIME
    LOG("Time: %dms\n", int((clock() - nol_cl) / (qwq)CLOCKS_PER_SEC * 1000));
#endif
    return ~~(0^_^0); /*さようならプログラム*/
}
    
int Ame__ = Ame_();
    
int main(){;}

标签：node,opt,P4848,rs,题解,BZOJ4605,ls,i32,x1
来源： https://www.cnblogs.com/Ame-sora/p/14425262.html