UVA - 10245 The Closest Pair Problem

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 cmp(struct coordination a,struct coordination b){
	if(a.x==b.x)return a.y<b.y;
	return a.x<b.x;
}
int compare(const void *x,const void *y){
	struct coordination* a=(struct coordination* )x;
	struct coordination* b=(struct coordination* )y;
	if(a->x==b->x)return a->y-b->y;
	return a->x-b->x;
}
void print(int n){
	for(int i=0;i<n;i++)
		cout<<a[i].x<<" "<<a[i].y<<endl;

}
double min(double a,double b){
	return a>b?b:a;
}
double getEucleanDistance(int i,int j){
	return sqrt(((a[i].x-a[j].x)*(a[i].x-a[j].x)+(a[i].y-a[j].y)*(a[i].y-a[j].y)));
}
double closestPair(int l,int r){
	if(l+1==r){
		return (numeric_limits<double>::max)();
	}
	int mid=(l+r)/2;
	double dl=closestPair(l,mid);
	double dr=closestPair(mid,r);
	double d=min(dl,dr);
	int count;
	for(int i=l;i<mid;i++){
		if(a[i].x>=a[mid].x-d){
			count=0;
			for(int j=mid;j<r&&count<6;j++){ //比较最多不超过6个点
				if(a[j].x<=a[mid].x+d&&a[j].y>=a[i].y-d&&a[j].y<=a[i].y+d){
					count++;
                    d=min(d,getEucleanDistance(i,j));
				}
			}
		}
	}
	return d;
}
int main() {
	int n;
	while(cin>>n,n){
		for(int i=0;i<n;i++)
			cin>>a[i].x>>a[i].y;
		sort(a,a+n,cmp);//sort t=158,while qsort t=169
		//qsort(a,n,sizeof(a[0]),compare);
		//print(n);
		//cout<<(numeric_limits<double>::max)()<<endl;
		double cd=closestPair(0,n);
		if(cd>=10000)cout<<"INFINITY\n";
		else
			printf("%.4f\n",cd);

	}
	return 0;
}