HDOJ-1007 Quoit Design（最近点对问题）

http://acm.hdu.edu.cn/showproblem.php?pid=1007

给出n个玩具（抽象为点）的坐标 求套圈的半径 要求最多只能套到一个玩具

实际就是要求最近的两个坐标的距离

典型的最近点对问题

 

最近点对详解

http://blog.csdn.net/lonelycatcher/article/details/7973046

//最近点对


# include <stdio.h>

# include <algorithm>

# include <math.h>

# define MAX 100005

# define INF 100000000

using namespace std;



struct POINT

{

	double x, y;

}arr[MAX];



int n, temp[MAX];



bool cmp(const POINT& a, const POINT& b)

{

	if(a.x != b.x)

		return a.x < b.x;

	return a.y < b.y;

}



bool cmp2(const int& a, const int& b)

{

	return arr[a].y < arr[b].y;

}



double dis(int a, int b)

{

	return sqrt(pow(arr[a].x - arr[b].x, 2) + pow(arr[a].y - arr[b].y, 2));

}



double min(double a, double b)

{

	return a<b?a:b;

}



double closest_pointpair(int left, int right)

{

	double d = INF;



	if(left == right)//单个点不计算直接返回INF

		return d;

	if(left + 1 == right)//相邻点直接计算距离

		return dis(left, right);



	int mid = (left + right) / 2;//取区间中值

	d = min(closest_pointpair(left, mid), closest_pointpair(mid+1, right));//分治 取二者中的小数作为领域范围



	int num = 0;

	for(int i = left; i <= right; i++)//记录中值两边+(-)d范围内的所有点的下标

		if(fabs(arr[mid].x - arr[i].x) <= d)

			temp[num++] = i;



	sort(temp, temp+num, cmp2);//按y从小到大排序



	for(int i = 0; i < num; i++)//两两枚举 若距离小于d 替换

	{

		for(int j = i + 1; j < num && arr[temp[j]].y - arr[temp[i]].y < d; j++)

		{

			double dt = dis(temp[i], temp[j]);

			d = min(d, dt);

		}

	}



	return d;//返回当前分域最近点对距离

}



int main()

{

	while(scanf("%d", &n) && n)

	{

		for(int i = 0; i < n; i++)

			scanf("%lf %lf",&arr[i].x, &arr[i].y);



		sort(arr, arr+n, cmp);



		printf("%.2lf\n", closest_pointpair(0, n-1) / 2);

	}



	return 0;		

}