ZOJ2107 Quoit Design 最近点对

ZOJ2107 给定10^5个点，求距离最近的点对的距离。 O（n^2）的算法是显而易见的。

可以通过分治优化到O（nlogn）

代码很简单

#include<iostream>
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<math.h>
#include<algorithm>
#include<queue>
#include<vector>
using namespace std;

const double eps=1e-9;

int cmp(double x)
{
 if(fabs(x)<eps)return 0;
 if(x>0)return 1;
 	else return -1;
}

const double pi=acos(-1.0);

inline double sqr(double x)
{
 return x*x;
}

struct point
{
 double x,y;
 point (){}
 point (double a,double b):x(a),y(b){}
 void input()
 	{
 	 scanf("%lf%lf",&x,&y);
	}
 friend point operator +(const point &a,const point &b)
 	{
 	 return point(a.x+b.x,a.y+b.y);
	}
 friend point operator -(const point &a,const point &b)
 	{
 	 return point(a.x-b.x,a.y-b.y);
	}
 friend bool operator ==(const point &a,const point &b)
 	{
 	 return cmp(a.x-b.x)==0&&cmp(a.y-b.y)==0;
	}
 friend point operator *(const point &a,const double &b)
 	{
 	 return point(a.x*b,a.y*b);
	}
 friend point operator*(const double &a,const point &b)
 	{
 	 return point(a*b.x,a*b.y);
	}
 friend point operator /(const point &a,const double &b)
 	{
 	 return point(a.x/b,a.y/b);
	}
 double norm()
 	{
 	 return sqrt(sqr(x)+sqr(y));
	}
};

const int maxn=100000+10;
point a[maxn];
int n,s[maxn];
bool cmpx(int i,int j)
{
 return cmp(a[i].x-a[j].x)<0;
}

bool cmpy(int i,int j)
{
 return cmp(a[i].y-a[j].y)<0;
}

double min_dist(point a[],int s[],int l,int r)
{
 double ans=1e100;
 if(r-l<20)
 	{
 	 for(int q=l;q<r;q++)
 	 	for(int w=q+1;w<r;w++)ans=min(ans,(a[s[q]]-a[s[w]]).norm());
 	 return ans;
	}
 int tl,tr,m=(l+r)/2;
 ans=min(min_dist(a,s,l,m),min_dist(a,s,m,r));
 for(tl=l;a[s[tl]].x<a[s[m]].x-ans;tl++);
 for(tr=r-1;a[s[tr]].x>a[s[m]].x+ans;tr--);
 sort(s+tl,s+tr,cmpy);
 for(int q=tl;q<tr;q++)
 	for(int w=q+1;w<min(tr,q+5);w++)
 		ans=min(ans,(a[s[q]]-a[s[w]]).norm());
 sort(s+tl,s+tr,cmpx);
 return ans;
}

double Min_Dist(point a[],int s[],int n)
{
 for(int i=0;i<n;++i)s[i]=i;
 sort(s,s+n,cmpx);
 return min_dist(a,s,0,n);
}

int main()
{freopen("t.txt","r",stdin);
 int n;
 while(scanf("%d",&n)&&n!=0)
 	{
 	 for(int i=0;i<n;i++)
 	 	a[i].input();
 	 printf("%.2lf\n",Min_Dist(a,s,n)/2.);
	}
 return 0;
}