BZOJ 2626 & KDtree

时间：2023-03-10 04:37:39
BZOJ 2626 & KDtree

题意:

　　二维平面n个点 每次给出一个点询问距离第k小的点.

SOL:

　　kdtree裸题,抄了一发别人的模板...二维割起来还是非常显然的.膜rzz的论文.

　　不多说了吧....

Code:

　　

/*==========================================================================

# Last modified: 2016-03-18 20:26

# Filename: 2626.cpp

# Description:

==========================================================================*/

#define me AcrossTheSky

#include <cstdio>

#include <cmath>

#include <ctime>

#include <string>

#include <cstring>

#include <cstdlib>

#include <iostream>

#include <algorithm> 

#include <set>

#include <map>

#include <stack>

#include <queue>

#include <vector> 

#define lowbit(x) (x)&(-x)

#define FOR(i,a,b) for((i)=(a);(i)<=(b);(i)++)

#define FORP(i,a,b) for(int i=(a);i<=(b);i++)

#define FORM(i,a,b) for(int i=(a);i>=(b);i--)

#define ls(a,b) (((a)+(b)) << 1)

#define rs(a,b) (((a)+(b)) >> 1)

#define getlc(a) ch[(a)][0]

#define getrc(a) ch[(a)][1] 

#define maxn 100000

#define maxm 100000

#define pi 3.1415926535898

#define _e 2.718281828459

#define INF 1070000000

using namespace std;

typedef long long LL;

typedef unsigned long long uLL; 

template<class T> inline

void read(T& num) {

    bool start=false,neg=false;

    char c;

    num=0; 

	while((c=getchar())!=EOF) {

        if(c=='-') start=neg=true;

        else if(c>='0' && c<='9') {

            start=true;

            num=num*10+c-'0';

        } else if(start) break;

    }

    if(neg) num=-num;

}

/*==================split line==================*/

int flag;

struct node{

 	int id;

   	LL x,y;

   	void in(int i){

   		read(x); read(y);

   		id=i;

   	}

   	bool operator<(const node& a)const{

   		if(!flag) return x<a.x;

   		return y<a.y;

   	}

}p[maxn];

LL sqr(LL x){return x*x;}

LL dist(node a,node b){return sqr(a.x-b.x)+sqr(a.y-b.y);}

struct cpt{

  	LL dis;

  	int id;

   	bool operator<(const cpt &a)const{

   		if(dis!=a.dis) return dis>a.dis;

   		return id<a.id;

   	}

};

priority_queue<cpt> q;	

struct KD{

   	LL minv[2],maxv[2];

   	LL dis(node p){

   		LL x=max(abs(p.x-minv[0]),abs(p.x-maxv[0]));

   		LL y=max(abs(p.y-minv[1]),abs(p.y-maxv[1]));

   		return sqr(x)+sqr(y);

  	}

   	KD operator+(const KD &a){

   		KD c;

   		c.minv[0]=min(minv[0],a.minv[0]);

   		c.minv[1]=min(minv[1],a.minv[1]);

   		c.maxv[0]=max(maxv[0],a.maxv[0]);

   		c.maxv[1]=max(maxv[1],a.maxv[1]);

   		return c;

   	}

};

struct KDtree{

    node p;KD c;

}tree[maxn];

int n,m,kth,ch[maxn][2],tot;

int build(int l,int r,int k){

	int x=++tot;

	flag=k;

	int mid=rs(l,r); nth_element(p+l,p+mid,p+r+1);

	tree[x].p=p[mid];

	tree[x].c.minv[0]=tree[x].c.maxv[0]=tree[x].p.x;

	tree[x].c.minv[1]=tree[x].c.maxv[1]=tree[x].p.y;

	if (l<=mid-1){

		ch[x][0]=build(l,mid-1,k^1);

		tree[x].c=tree[x].c+tree[ch[x][0]].c;

	}

	if (mid+1<=r){

		ch[x][1]=build(mid+1,r,k^1);

		tree[x].c=tree[x].c+tree[ch[x][1]].c;

	}

	return x;

}

void ask(int x,node a,int k){

	cpt tmp=(cpt){dist(tree[x].p,a),tree[x].p.id};

	int lc=ch[x][0],rc=ch[x][1];

	flag=k;

	if (q.size()<kth) q.push(tmp);

	else if (tmp<q.top()){q.pop(); q.push(tmp);}

	if (a<tree[x].p) swap(lc,rc);

	if (lc && (q.size()<kth || tree[lc].c.dis(a)>=q.top().dis))

		ask(lc,a,k^1);

	if (rc && (q.size()<kth || tree[rc].c.dis(a)>=q.top().dis))

		ask(rc,a,k^1);

}

int main(){

	//freopen("a.in","r",stdin);

	read(n);

	FORP(i,1,n) p[i].in(i);

	build(1,n,0);

	int m; read(m);

	FORP(i,1,m){

		node a; a.in(i);

		read(kth);

		while (!q.empty()) q.pop();

		ask(1,a,0);

		printf("%d\n",q.top().id);

	}

}

相关文章