题意:求给出的所有点的最远点减最近点的最小差值
KD树的最远估价和最近估价略微不同,直接找最远垂线,反正xjb改一下就过了
#include<bits/stdc++.h>
#define rep(i,j,k) for(register int i=j;i<=k;i++)
#define rrep(i,j,k) for(register int i=j;i>=k;i--)
#define erep(i,u) for(register int i=head[u];~i;i=nxt[i])
#define print(a) printf("%lld",(ll)(a))
#define println(a) printf("%lld\n",(ll)(a))
#define printbk(a) printf("%lld ",(ll)(a))
using namespace std;
const int MAXN = 5e5+11;
const int INF = 0x7fffffff;
typedef long long ll;
ll read(){
ll x=0,f=1;register char ch=getchar();
while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
return x*f;
}
int D;
struct point{
int x[2];
bool operator < (const point &rhs) const{
return x[D]<rhs.x[D];
}
};
struct KD{
int son[MAXN][2];
point p[MAXN],mn[MAXN],mx[MAXN];
int root,ans,tot;
void init(){
ans=INF; tot=D=0;
}
void pu(int o){
rep(i,0,1){
if(son[o][i]) rep(j,0,1){
if(mn[son[o][i]].x[j]<mn[o].x[j]) mn[o].x[j]=mn[son[o][i]].x[j];
if(mx[son[o][i]].x[j]>mx[o].x[j]) mx[o].x[j]=mx[son[o][i]].x[j];
}
}
}
int build(int now,int l,int r){
int mid=l+r>>1;
tot++; son[mid][0]=son[mid][1]=0;
D=now;nth_element(p+l,p+mid,p+r+1);//[l,r+1)
mn[mid].x[0]=mx[mid].x[0]=p[mid].x[0];
mn[mid].x[1]=mx[mid].x[1]=p[mid].x[1];
if(l<mid) son[mid][0]=build(now^1,l,mid-1);
if(r>mid) son[mid][1]=build(now^1,mid+1,r);
pu(mid);
return mid;
}
void insert(int &o,int now,point v){
if(!o){
o=++tot;
p[o].x[0]=mn[o].x[0]=mx[o].x[0]=v.x[0];
p[o].x[1]=mn[o].x[1]=mx[o].x[1]=v.x[1];
}else{
insert(son[o][p[o].x[now]<v.x[now]],now^1,v);
pu(o);
}
}
inline int dis(point &a,point &b){
return abs(a.x[0]-b.x[0])+abs(a.x[1]-b.x[1]);
}
inline int eva(int o,point &v){
int res=0;
rep(i,0,1) if(v.x[i]<mn[o].x[i]||v.x[i]>mx[o].x[i]){
if(v.x[i]<mn[o].x[i]) res+=mn[o].x[i]-v.x[i];
else res+=v.x[i]-mx[o].x[i];
}
return res;
}
void query1(int o,point v){
if(!o) return;
int d1=dis(p[o],v),d2=INF,d3=INF;
if(d1<ans&&d1!=0) ans=d1;
if(son[o][0]) d2=eva(son[o][0],v);
if(son[o][1]) d3=eva(son[o][1],v);
if(d2<d3){
if(d2<ans) query1(son[o][0],v);
if(d3<ans) query1(son[o][1],v);
}else{
if(d3<ans) query1(son[o][1],v);
if(d2<ans) query1(son[o][0],v);
}
}
inline int eva2(int o,point &v){
int res=0;
rep(i,0,1) res+=max(abs(v.x[i]-mn[o].x[i]),abs(v.x[i]-mx[o].x[i]));
return res;
}
void query2(int o,point v){
if(!o) return;
int d1=dis(p[o],v),d2=0,d3=0;
if(d1>ans) ans=d1;
if(son[o][0]) d2=eva2(son[o][0],v);
if(son[o][1]) d3=eva2(son[o][1],v);
if(d2>d3){
if(d2>ans) query2(son[o][0],v);
if(d3>ans) query2(son[o][1],v);
}else{
if(d3>ans) query2(son[o][1],v);
if(d2>ans) query2(son[o][0],v);
}
}
inline ll queryMin(point v){
ans=INF;
query1(root,v);
return ans;
}
inline ll queryMax(point v){
ans=0;
query2(root,v);
return ans;
}
}kd;
int main(){
int n;
while(cin>>n){
kd.init();
rep(i,1,n){
kd.p[i].x[0]=read();
kd.p[i].x[1]=read();
}
kd.root=kd.build(0,1,n);
ll res=INF;
rep(i,1,n){
res=min(res,kd.queryMax(kd.p[i])-kd.queryMin(kd.p[i]));
}
println(res);
}
return 0;
}