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;
}