hdu 1007 Quoit Design(分治法求最近点对)

大致题意：给N个点，求最近点对的距离 d ；输出：r = d/2。

// Time 2093 ms; Memory 1812 K

#include<iostream>
#include<cstdio>
#include<cmath>
#include<algorithm>
#define eps 1e-8
#define maxn 100010
#define sqr(a) ((a)*(a))

using namespace std;

int sig(double x)
{
	return (x>eps)-(x<-eps);
}

struct point
{
	double x,y;
	point(double xx=0,double yy=0):x(xx),y(yy){}
}p[maxn];

bool operator < (point a,point b)
{
	return a.x<b.x || (a.x==b.x && a.y<b.y);
}
bool cmp(point a,point b)
{
	return a.y<b.y || (a.y==b.y && a.x<b.x);
}
double len(point a,point b)
{
	return sqrt(sqr(a.x-b.x)+sqr(a.y-b.y));
}
double min_dis(int l,int r)
{
	int i,j,k,s;
	double d,mi=-1;
	if(r-l<4)
	{
		for(i=l;i<r;i++) for(j=i+1;j<r;j++)
		{
			d=len(p[i],p[j]);
			if(mi<0 || sig(d-mi)<0) mi=d;
		}
		return mi;
	}
	int mid=(l+r)/2;
	double minl=min_dis(l,mid);
	double minr=min_dis(mid,r);
	mi=minl<minr?minl:minr;
	for(i=1;i<mid-l && sig(p[mid].x-p[mid-i].x-mi)<0;i++);i--;
	for(j=1;j<r-mid && sig(p[mid+j].x-p[mid].x-mi)<0;j++);j--;
	sort(p+mid-i,p+mid,cmp);
	sort(p+mid,p+mid+j+1,cmp);
	int t=mid,flag;
	for(k=mid-i;k<mid;k++)
	{
		flag=1;
		for(s=t;s<=mid+j;s++)
		{
			if(sig(p[k].y-p[s].y)>0)
			{
				if(sig(p[k].y-p[s].y-mi)>=0) continue;
				else
				{
					if(flag)
					{
						flag=0;t=s;
					}
					d=len(p[s],p[k]);
					if(sig(d-mi)<0) mi=d;
				}
			}
			else
			{
				if(sig(p[s].y-p[k].y-mi)>=0) break;
				else
				{
					d=len(p[s],p[k]);
					if(sig(d-mi)<0) mi=d;
				}
			}
		}
	}
	sort(p+mid-i,p+mid+j+1);
	return mi;
}

int main()
{
	int i,n;
	while(scanf("%d",&n)!=EOF && n)
	{
		for(i=0;i<n;i++) scanf("%lf%lf",&p[i].x,&p[i].y);
		sort(p,p+n);
		printf("%.2lf\n",min_dis(0,n)/2);
	}
	return 0;
}