[KD-TREE] BZOJ 1941 [Sdoi2010]Hide and Seek

题意：曼哈顿距离最大值与最小值的差

模板题

#include
#include
#include
using namespace std;

inline char nc()
{
	static char buf[100000],*p1=buf,*p2=buf;
	if (p1==p2) { p2=(p1=buf)+fread(buf,1,100000,stdin); if (p1==p2) return EOF; }
	return *p1++;
}

inline void read(int &x)
{
	char c=nc(),b=1;
	for (;!(c>='0' && c<='9');c=nc()) if (c=='-') b=-1;
	for (x=0;c>='0' && c<='9';x=x*10+c-'0',c=nc()); x*=b;
}

const int N=500005;

int n,D;

struct node{
	int d[2],mx[2],mn[2],l,r;
	node(int x=0,int y=0) { mx[0]=mn[0]=d[0]=x; mx[1]=mn[1]=d[1]=y; l=r=0; } 
	int& operator[](int x) { return d[x]; }
	friend bool operator == (const node &A, const node &B) { return A.d[0]==B.d[0] && A.d[1]==B.d[1]; }
	friend bool operator < (const node &A,const node &B) { return A.d[D]r) return 0;
		int mid=(l+r)>>1;
		::D=D; nth_element(tmp+l,tmp+mid,tmp+r+1);
		T[mid]=tmp[mid];
		T[mid].l=Reb(l,mid-1,D^1);
		T[mid].r=Reb(mid+1,r,D^1);
		update(mid); return mid;
	}
	int Get_mn(node A){
		int ret=0;
		for(int i=0;i<2;i++)
			ret+=max(now[i]-A.mx[i],0),ret+=max(A.mn[i]-now[i],0);
		return ret;
	}
	int Que_mn(int x){
		int tmp=dist(T[x],now);
		if (tmp) ans=min(ans,tmp);
		int l=T[x].l,r=T[x].r,dl=1<<30,dr=1<<30;
		if(l) dl=Get_mn(T[l]); 
		if(r) dr=Get_mn(T[r]);
		if(dldr){
			if(dl>ans) Que_mx(l);
			if(dr>ans) Que_mx(r);
		}else {
			if(dr>ans) Que_mx(r);
			if(dl>ans) Que_mx(l);
		}
		return ans;
	}
}KD;

int x[N],y[N];
int ans=1<<30;

int main()
{
	int minimum,maximum;
	freopen("t.in","r",stdin);
	freopen("t.out","w",stdout);
	read(n);
	for (int i=1;i<=n;i++)
		read(x[i]),read(y[i]),tmp[i]=node(x[i],y[i]);
	KD.root=KD.Reb(1,n,0);
	for (int i=1;i<=n;i++)
	{
		KD.now=node(x[i],y[i]);
		KD.ans=-1<<30; maximum=KD.Que_mx(KD.root);
		KD.ans= 1<<30; minimum=KD.Que_mn(KD.root);
		ans=min(ans,maximum-minimum);
	}
	printf("%d\n",ans);
	return 0;
}