BZOJ 2648 SJY摆棋子 K-Dimensional-Tree

标签：bzoj   bzoj2648   k-dimensional-tree   

题目大意：给定平面上的n个点，定义距离为曼哈顿距离，支持下列操作：

1.插入一个点

2.查询离一个点最近的点的距离

Hint说KDTree【可以】过，那么不写KDT还能写啥= =

我的CDQ分治可是T掉了啊= =

记住KDT发生TLE事件的时候不一定是常数问题 有可能写挂了= =(这不和莫队一样么233

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#define M 500500
#define INF 0x3f3f3f3f
using namespace std;
struct Point{
	int x,y;
	friend istream& operator >> (istream &_,Point &p)
	{
		scanf("%d%d",&p.x,&p.y);
		return _;
	}
	friend int Distance(const Point &p1,const Point &p2)
	{
		return abs(p1.x-p2.x) + abs(p1.y-p2.y) ;
	}
}points[M];
int n,m;
bool Compare_x(const Point &p1,const Point &p2)
{
	return p1.x < p2.x ;
}
bool Compare_y(const Point &p1,const Point &p2)
{
	return p1.y < p2.y ;
}
namespace K_Dimensional_Tree{
	struct abcd{
		abcd *ls,*rs;
		Point p;
		int x1,y1,x2,y2;
		abcd() {}
		abcd(const Point &_):p(_)
		{
			ls=rs=0x0;
			x1=x2=p.x;
			y1=y2=p.y;
		}
		void Push_Up(abcd *x)
		{
			x1=min(x1,x->x1);
			y1=min(y1,x->y1);
			x2=max(x2,x->x2);
			y2=max(y2,x->y2);
		}
		int Min_Distance(const Point &p)
		{
			int re=0;
			if(p.x<x1) re+=x1-p.x;
			if(p.x>x2) re+=p.x-x2;
			if(p.y<y1) re+=y1-p.y;
			if(p.y>y2) re+=p.y-y2;
			return re;
		}
		/*
		int Max_Distance(const Point &p)
		{
			int re=0;
			re+=max(p.x-x1,x2-p.x);
			re+=max(p.y-y1,y2-p.y);
			return re;
		}
		*/
	}mempool[M],*C=mempool,*root;
	void Build_Tree(abcd *&x,int l,int r,bool flag)
	{
		if(l>r)
			return ;
		int mid=l+r>>1;
		nth_element(points+l,points+mid,points+r+1,flag?Compare_y:Compare_x);
		x=new abcd(points[mid]);
		Build_Tree(x->ls,l,mid-1,flag^1);
		Build_Tree(x->rs,mid+1,r,flag^1);
		if(x->ls) x->Push_Up(x->ls);
		if(x->rs) x->Push_Up(x->rs);
	}
	void Insert(abcd *&x,const Point &p,bool flag)
	{
		if(!x)
		{
			x=new abcd(p);
			return ;
		}
		if( (flag?Compare_y:Compare_x)(p,x->p) )
		{
			Insert(x->ls,p,flag^1);
			x->Push_Up(x->ls);
		}
		else
		{
			Insert(x->rs,p,flag^1);
			x->Push_Up(x->rs);
		}
	}
	void Get_Min(abcd *x,const Point &p,int &ans)
	{
		ans=min(ans,Distance(x->p,p));
		int l_dis=x->ls?x->ls->Min_Distance(p):INF;
		int r_dis=x->rs?x->rs->Min_Distance(p):INF;
		if(l_dis<r_dis)
		{
			if( x->ls && l_dis<ans )
				Get_Min(x->ls,p,ans);
			if( x->rs && r_dis<ans )
				Get_Min(x->rs,p,ans);
		}
		else
		{
			if( x->rs && r_dis<ans )
				Get_Min(x->rs,p,ans);
			if( x->ls && l_dis<ans )
				Get_Min(x->ls,p,ans);
		}
	}
}
int main()
{
	using namespace K_Dimensional_Tree;
	int i,o;
	Point p;
	cin>>n>>m;
	for(i=1;i<=n;i++)
		cin>>points[i];
	Build_Tree(root,1,n,0);
	for(i=1;i<=m;i++)
	{
		scanf("%d",&o);cin>>p;
		if(o==1)
			Insert(root,p,0);
		else
		{
			int ans=INF;
			Get_Min(root,p,ans);
			printf("%d\n",ans);
		}
	}
	return 0;
}

BZOJ 2648 SJY摆棋子 K-Dimensional-Tree

标签：bzoj   bzoj2648   k-dimensional-tree   

原文地址：http://blog.csdn.net/popoqqq/article/details/45056709