【HDU 1007】 Quoit Design

#include <algorithm>  
#include <bitset>  
#include <cctype>  
#include <cerrno>  
#include <clocale>  
#include <cmath>  
#include <complex>  
#include <cstdio>  
#include <cstdlib>  
#include <cstring>  
#include <ctime>  
#include <deque>  
#include <exception>  
#include <fstream>  
#include <functional>  
#include <limits>  
#include <list>  
#include <map>  
#include <iomanip>  
#include <ios>  
#include <iosfwd>  
#include <iostream>  
#include <istream>  
#include <ostream>  
#include <queue>  
#include <set>  
#include <sstream>  
#include <stdexcept>  
#include <streambuf>  
#include <string>  
#include <utility>  
#include <vector>  
#include <cwchar>  
#include <cwctype>  
#include <stack>  
#include <limits.h> 
using namespace std;
#define MAXN 100010
const double INF = 1e10;

struct info
{
        double x,y;        
} a[MAXN];

int n,i;

inline bool cmpx(info a,info b) 
{
        return a.x < b.x;
}
inline bool cmpy(info a,info b)
{
        return a.y < b.y;
}
inline double dist(info p,info q)
{
        return sqrt(abs(p.x - q.x) * abs(p.x - q.x) + abs(p.y - q.y) * abs(p.y - q.y));
}
inline double Closest_Pair(int l,int r)
{
        int i,j,mid,len = 0;
        static info s[MAXN];
        double d;
        if (l == r) return INF; 
        if (l + 1 == r) return dist(a[l],a[r]);
        mid = (l + r) >> 1;
        d = min(Closest_Pair(l,mid),Closest_Pair(mid+1,r));
        for (i = l; i <= r; i++)
        {
                if (abs(a[mid].x - a[i].x) <= d) s[++len] = a[i];    
        }        
        sort(s+1,s+len+1,cmpy);
        for (i = 1; i <= len; i++)
        {
                for (j = i + 1; j <= len && s[j].y - s[i].y <= d; j++)
                {
                        d = min(d,dist(s[i],s[j]));
                }
        }
        return d;
}

int main() 
{
        
        while (scanf("%d",&n) && n)
        {
                for (i = 1; i <= n; i++) scanf("%lf%lf",&a[i].x,&a[i].y);
                sort(a+1,a+n+1,cmpx);
                printf("%.2lf\n",Closest_Pair(1,n) / 2.0);
        }
        
        return 0;
    
}