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.****.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;
}