为什么要写这个题、、经典啊，当然，别以为我用分治做的，不过主要思想还是那神奇的六个点共存（一个h*2h的矩形中最多能放下多少个点使得两两距离不超过h）
其实我是在这里看到的
http://community.topcoder.com/tc?module=Static&d1=tutorials&d2=lineSweep
排个序，然后扫描过去，每次确定Y的范围，暴力找每个点（其实这是O(1)的）
蛮不错的哦。
写完后发现比分治写的快了好多啊，估计是我不会写分治吧T_T，总之，现在跑到了rank 4了。。。
数据结构就是这样的直接啊！
/* **********************************************
Author : *qi
Created Time: 2013-8-24 12:54:35
File Name : ruocai.cpp
*********************************************** */
#include <cstdio>
#include <cmath>
#include <stack>
#include <algorithm>
using std::stack;
using std::pair;
using std::make_pair;
#define L x->c[0]
#define R x->c[1]
#define KT root->c[1]->c[0]
const int maxn = 200010;
struct node{
node *c[2] , *fa;
int id,sz;
double val,x;
inline bool d() {
return fa->c[0] == this;
}
inline void setc(int d,node *s) {
c[d] = s;
s->fa = this;
}
inline void up() {
sz = c[0]->sz + c[1]->sz + 1;
}
}NODE[maxn],*null=&NODE[0];
node *ID[maxn];
int top;
struct Splay_tree{
node *root;
void Rotate(node *x,int f){
node *y = x->fa;
y->setc(!f,x->c[f]);
x->fa = y->fa;
if(y->fa != null) y->fa->setc(!y->d(),x);
x->setc(f,y);
y->up();
}
void Splay(node *x,node *goal) {
while(x->fa!=goal) {
if(x->fa->fa == goal) Rotate(x,x->d());
else {
int f = x->fa->d();
x->d() == f ? Rotate(x->fa,f) : Rotate(x,!f);
Rotate(x,f);
}
}
x->up();
if(goal == null) {
root = x;
}
}
void RTO(int k,node *goal) {
node *x = root;
while(L->sz + 1 != k) {
if(k < L->sz + 1) x = L;
else {
k -= L->sz + 1;
x = R;
}
}
Splay(x,goal);
}
void insert(node* &x,node *y) {
if(x == null) {
x = y;
return ;
}
if(y->val <= x->val) {
insert(x->c[0],y);
x->c[0]->fa = x;
} else {
insert(x->c[1],y);
x->c[1]->fa = x;
}
x->up();
}
node* new_node(int id,double x,double y) {
node* tmp = &NODE[++top];
tmp->fa = null;
tmp->val = y;
tmp->x = x;
tmp->sz = 1;
tmp->id = top;
tmp->c[0] = tmp->c[1] = null;
ID[id] = tmp;
return tmp;
}
void insert(int id,double x,double y) {
node *tmp = new_node(id,x,y);
insert(root,tmp);
}
void init() {
root = null;
}
void Del_root() {
node *t = root;
if(t->c[1] != null) {
root = t->c[1];
RTO(1,null);
root->c[0] = t->c[0];
if(root->c[0] != null)
root->c[0]->fa = root;
} else {
root = root->c[0];
}
root->fa = null;
if(root != null) root->up();
} void Del(node *tmp) {
Splay(tmp,null);
Del_root();
}
node* find_pre(node *x,double v) {
if(x == null) return null;
if(x->val < v) {
node *tmp = find_pre(x->c[1],v);
return tmp == null ? x : tmp;
}
else {
return find_pre(x->c[0],v);
}
}
node* find_succ(node *x,double v) {
if(x == null) return null;
if(x->val > v ) {
node *tmp = find_succ(x->c[0],v);
return tmp == null ? x : tmp;
} else {
return find_succ(x->c[1],v);
}
}
void search(double a,double b,double &ans,node* x){
if(x == null) return ;
double c = x->x , d = x->val;
double cost = sqrt((a-c)*(a-c) + (b-d)*(b-d));
if(cost < ans) ans = cost;
if(x->c[0] != null) search(a,b,ans,x->c[0]);
if(x->c[1] != null) search(a,b,ans,x->c[1]);
}
void gao(double x,double y,double &ans) {
node *pre = find_pre(root,y-ans) ;
node *succ = find_succ(root,y+ans);
if(pre == null || succ == null) while(1);
Splay(pre,null);
Splay(succ,root);
search(x,y,ans,KT);
}
void vist(node *x) {
if(x!=null) {
printf("%d x = %lf y = %lf lson=%d rson=%d\n",x->id,x->x,x->val,x->c[0]->id,x->c[1]->id);
if(x->c[0]!=null) vist(x->c[0]);
if(x->c[1]!=null) vist(x->c[1]);
}
}
void debug(){
puts("*******");
vist(root);
puts("*****");
}
}spt;
void prepare() {
null->id = 0;
null->c[0] = null->c[1] = null->fa = NULL;
null->sz = null->val = 0;
top = 0;
}
double x[maxn] , y[maxn];
bool by_x(int a,int b) {
return x[a] > x[b];
}
int id[maxn];
double sqr(double a) { return a * a; }
int main()
{
int n;
while(scanf("%d",&n),n) {
prepare();
spt.init();
for(int i = 0; i < n; i++) {
id[i] = i;
scanf("%lf%lf",&x[i],&y[i]);
}
if(n == 1){
puts("0.00");
continue;
}
std::sort(id,id+n,by_x);
spt.insert(0,-1e100,-1e100);
spt.insert(0,-1e100,1e100);
double ans = 1e50 ;
int pt = 0;
for(int i = 0; i < n; i++) {
if(ans == 0) break;
while(pt < i && x[id[pt]] > x[id[i]] + ans ) {
if(spt.root->sz == 2) break;
spt.Del(ID[id[pt]]);
pt++;
}
if(spt.root->sz == 2) {
spt.insert(id[i],x[id[i]],y[id[i]]);
} else {
spt.gao(x[id[i]],y[id[i]],ans);
spt.insert(id[i],x[id[i]],y[id[i]]);
}
}
printf("%.2f\n",ans*0.5);
}
return 0;
}