题目大意：
n个点，求每个点到其最远点距离-到其最近点距离（除自己之外）的最小值
思路：
对于估计函数的理解还不够深刻
1 #include<iostream> 2 #include<cstdio> 3 #include<cstring> 4 #include<cstdlib> 5 #include<cmath> 6 #include<algorithm> 7 #include<queue> 8 #include<vector> 9 #include<map> 10 #include<set> 11 #define ll long long 12 #define inf 2139062143 13 #define MAXN 500100 14 #define MOD 998244353 15 #define rep(i,s,t) for(register int i=(s),i##__end=(t);i<=i##__end;++i) 16 #define dwn(i,s,t) for(register int i=(s),i##__end=(t);i>=i##__end;--i) 17 #define ren(x) for(register int i=fst[x];i;i=nxt[i]) 18 #define pb(i,x) vec[i].push_back(x) 19 #define pls(a,b) (a+b)%MOD 20 #define mns(a,b) (a-b+MOD)%MOD 21 #define mul(a,b) (1LL*(a)*(b))%MOD 22 using namespace std; 23 inline int read() 24 { 25 int x=0,f=1;char ch=getchar(); 26 while(!isdigit(ch)) {if(ch=='-') f=-1;ch=getchar();} 27 while(isdigit(ch)) {x=x*10+ch-'0';ch=getchar();} 28 return x*f; 29 } 30 int rt,n,Dim,ret1,ret2,ans; 31 struct node{int mx[2],mn[2],d[2],l,r;}tr[MAXN],g; 32 bool operator < (const node a,const node b) {return a.d[Dim]<b.d[Dim];} 33 struct kd_tree 34 { 35 void upd(int k) 36 { 37 node gl=tr[tr[k].l],gr=tr[tr[k].r]; 38 rep(i,0,1) 39 { 40 if(tr[k].l) tr[k].mn[i]=min(tr[k].mn[i],gl.mn[i]),tr[k].mx[i]=max(tr[k].mx[i],gl.mx[i]); 41 if(tr[k].r) tr[k].mn[i]=min(tr[k].mn[i],gr.mn[i]),tr[k].mx[i]=max(tr[k].mx[i],gr.mx[i]); 42 } 43 } 44 int build(int l,int r,int D) 45 { 46 int mid=l+r>>1;Dim=D;nth_element(tr+l,tr+mid,tr+r+1); 47 rep(i,0,1) tr[mid].mn[i]=tr[mid].mx[i]=tr[mid].d[i]; 48 if(l<mid) tr[mid].l=build(l,mid-1,D^1); 49 if(mid<r) tr[mid].r=build(mid+1,r,D^1); 50 upd(mid);return mid; 51 } 52 int Get1(int k,int res=0) 53 { 54 rep(i,0,1) res+=max(abs(g.d[i]-tr[k].mn[i]),abs(g.d[i]-tr[k].mx[i])); 55 return res; 56 } 57 int Get2(int k,int res=0) 58 { 59 rep(i,0,1) res+=max(0,tr[k].mn[i]-g.d[i]); 60 rep(i,0,1) res+=max(0,g.d[i]-tr[k].mx[i]);return res; 61 } 62 int dis(int k,int res=0) {rep(i,0,1) res+=abs(tr[k].d[i]-g.d[i]);return res;} 63 void query(int k) 64 { 65 int dl1=0,dr1=0,dl2=inf,dr2=inf; 66 ret1=max(ret1,dis(k));if(dis(k)) ret2=min(ret2,dis(k)); 67 if(tr[k].l) dl1=Get1(tr[k].l),dl2=Get2(tr[k].l); 68 if(tr[k].r) dr1=Get1(tr[k].r),dr2=Get2(tr[k].r); 69 if(dl1>dr1) 70 { 71 if(dl1>ret1) 72 { 73 query(tr[k].l); 74 if(dr1>ret1||dr2<ret2) query(tr[k].r); 75 }else 76 { 77 if(dr2<ret2) query(tr[k].r); 78 if(dl2<ret2) query(tr[k].l); 79 } 80 }else 81 { 82 if(dr1>ret1) 83 { 84 query(tr[k].r); 85 if(dl1>ret1||dl2<ret2) query(tr[k].l); 86 }else 87 { 88 if(dl2<ret2) query(tr[k].l); 89 if(dr2<ret2) query(tr[k].r); 90 } 91 } 92 } 93 }kd; 94 int main() 95 { 96 n=read();rep(i,1,n) tr[i].d[0]=read(),tr[i].d[1]=read(); 97 rt=kd.build(1,n,0),ans=inf; 98 rep(i,1,n) 99 { 100 g=tr[i],ret1=0,ret2=inf;kd.query(rt); 101 ans=min(ans,ret1-ret2); 102 } 103 printf("%d ",ans); 104 }