UVA

属于Divide-and-Conquer，算法课老师有讲到，就找个题目试试，思想就是不断的二分。。

。考虑合并时的处理。。不解释

//============================================================================
// Name        : uva10245.cpp
// Author      : 
// Version     :
// Copyright   : Your copyright notice
// Description : Uva10245 in C++, Ansi-style
//============================================================================

#include <iostream>
#include<stdlib.h>
#include<stdio.h>
#include<math.h>
#include <limits>
#include<algorithm>

using namespace std;
const int N=10005;
struct coordination{
	double x,y;
}a[N];
int cmpstruct coordination a,struct coordination b){
	ifa.x==b.x)return a.y<b.y;
	return a.x<b.x;
}
int compareconst void *x,const void *y){
	struct coordination* a=struct coordination* )x;
	struct coordination* b=struct coordination* )y;
	ifa->x==b->x)return a->y-b->y;
	return a->x-b->x;
}
void printint n){
	forint i=0;i<n;i++)
		cout<<a[i].x<<" "<<a[i].y<<endl;

}
double mindouble a,double b){
	return a>b?
b:a;
}
double getEucleanDistanceint i,int j){
	return sqrta[i].x-a[j].x)*a[i].x-a[j].x)+a[i].y-a[j].y)*a[i].y-a[j].y)));
}
double closestPairint l,int r){
	ifl+1==r){
		return numeric_limits<double>::max));
	}
	int mid=l+r)/2;
	double dl=closestPairl,mid);
	double dr=closestPairmid,r);
	double d=mindl,dr);
	int count;
	forint i=l;i<mid;i++){
		ifa[i].x>=a[mid].x-d){
			count=0;
			forint j=mid;j<r&&count<6;j++){ //比較最多不超过6个点
				ifa[j].x<=a[mid].x+d&&a[j].y>=a[i].y-d&&a[j].y<=a[i].y+d){
					count++;
                    d=mind,getEucleanDistancei,j));
				}
			}
		}
	}
	return d;
}
int main) {
	int n;
	whilecin>>n,n){
		forint i=0;i<n;i++)
			cin>>a[i].x>>a[i].y;
		sorta,a+n,cmp);//sort t=158,while qsort t=169
		//qsorta,n,sizeofa[0]),compare);
		//printn);
		//cout<<numeric_limits<double>::max))<<endl;
		double cd=closestPair0,n);
		ifcd>=10000)cout<<"INFINITY
";
		else
			printf"%.4f
",cd);

	}
	return 0;
}