bzoj 1941: [Sdoi2010]Hide and Seek k-D Tree

bzoj 1941: [Sdoi2010]Hide and Seek

题目大意:

给n个点,找出一个点使到这个点到其他点的最大曼哈顿距离与最小曼哈顿距离之差最小。

题解

我们可以分别枚举每个点
然后对于该点查询到该点的最远点和最近点
这个直接用K-D Tree就好了嘛..

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
typedef long long ll;
inline void read(int &x){
	x=0;char ch;bool flag = false;
	while(ch=getchar(),ch<'!');if(ch == '-') ch=getchar(),flag = true;
	while(x=10*x+ch-'0',ch=getchar(),ch>'!');if(flag) x=-x;
}
inline int max(int a,int b,int c){return max(a,max(b,c));}
inline int min(int a,int b,int c){return min(a,min(b,c));}
const int maxn = 500010;
const int dem = 2;
struct Node{
	Node *ch[2];
	int pos[2],id;
	int minn[2],maxx[2];
	inline void update(){
		minn[0] = min(ch[0]->minn[0],ch[1]->minn[0],pos[0]);
		maxx[0] = max(ch[0]->maxx[0],ch[1]->maxx[0],pos[0]);
		minn[1] = min(ch[0]->minn[1],ch[1]->minn[1],pos[1]);
		maxx[1] = max(ch[0]->maxx[1],ch[1]->maxx[1],pos[1]);
	}
}*null,*root;
Node T[maxn],asd;
inline void init(){
	null = &asd;null->ch[0] = null->ch[1] = null;
	null->minn[0] = null->minn[1] = 0x3f3f3f3f;
	null->maxx[0] = null->maxx[1] = -0x3f3f3f3f;
}int nowd;
inline bool cmp(const Node &a,const Node &b){
	return a.pos[nowd] < b.pos[nowd];
}
Node *build(int l,int r,int s){
	if(l > r) return null;
	int mid = (l+r) >> 1;
	nowd = s % dem;
	nth_element(T+l,T+mid,T+r+1,cmp);
	Node *p = &T[mid];
	p->ch[0] = build(l,mid-1,s+1);
	p->ch[1] = build(mid+1,r,s+1);
	p->update();return p;
}
ll ans_max,ans_min;
inline ll cat_abs(int x){return x < 0 ? -x : x;}
Node op;
inline ll get_max(Node *p){
	ll ret = 0;
	ret = max(ret,cat_abs(op.pos[0]-p->minn[0]) + cat_abs(op.pos[1]-p->minn[1]));
	ret = max(ret,cat_abs(op.pos[0]-p->minn[0]) + cat_abs(op.pos[1]-p->maxx[1]));
	ret = max(ret,cat_abs(op.pos[0]-p->maxx[0]) + cat_abs(op.pos[1]-p->minn[1]));
	ret = max(ret,cat_abs(op.pos[0]-p->maxx[0]) + cat_abs(op.pos[1]-p->maxx[1]));
	return ret;
}
inline ll get_min(Node *p){
	ll ret = 0;
	for(int d=0;d<dem;d++) if(op.pos[d]<p->minn[d]||op.pos[d]>p->maxx[d])
		ret+=(op.pos[d]<p->minn[d])?p->minn[d]-op.pos[d]:op.pos[d]-p->maxx[d];
	return ret;
}
void query_max(Node *p){
	if(p == null) return;
	ll dis = 0;
	for(int d=0;d<dem;++d) dis += cat_abs(p->pos[d] - op.pos[d]);
	if(p->id != op.id) ans_max = max(ans_max,dis);
	if(get_max(p->ch[0]) < get_max(p->ch[1])) swap(p->ch[0],p->ch[1]);
	if(get_max(p->ch[0]) > ans_max)  query_max(p->ch[0]);
	if(get_max(p->ch[1]) > ans_max)  query_max(p->ch[1]);
}
void query_min(Node *p){
	if(p == null) return;
	ll dis = 0;
	for(int d=0;d<dem;++d) dis += cat_abs(p->pos[d] - op.pos[d]);
	if(p->id != op.id) ans_min = min(ans_min,dis);
	if(get_min(p->ch[0]) > get_min(p->ch[1])) swap(p->ch[0],p->ch[1]);
	if(get_min(p->ch[0]) < ans_min) query_min(p->ch[0]);
	if(get_min(p->ch[1]) < ans_min) query_min(p->ch[1]);
}
int main(){
	init();
	int n;read(n);
	for(int i=1;i<=n;++i){
		read(T[i].pos[0]);
		read(T[i].pos[1]);
		T[i].id = i;T[i].ch[0] = T[i].ch[1] = null;
		T[i].update();
	}root = build(1,n,1);
	ll ans = 1LL<<60;
	for(int i=1;i<=n;++i){
		op = T[i];
		ans_max = 0;
		ans_min = 1LL<<60;
		query_min(root);
		query_max(root);
		ans = min(ans,ans_max - ans_min);
	}printf("%lld\n",ans);
	getchar();getchar();
	return 0;
}