codeforce #514 D. Nature Reserve

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探讨如何在森林中设计一个圆形保护区，以最小半径覆盖所有动物栖息地并仅与河流相切一次。解决方法涉及坐标转换、二分查找及判断圆区间交集，确保保护区既保护动物又不影响河流交通。

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D. Nature Reserve

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

There is a forest that we model as a plane and live nn rare animals. Animal number ii has its lair in the point (xi,yi)(xi,yi). In order to protect them, a decision to build a nature reserve has been made.

The reserve must have a form of a circle containing all lairs. There is also a straight river flowing through the forest. All animals drink from this river, therefore it must have at least one common point with the reserve. On the other hand, ships constantly sail along the river, so the reserve must not have more than one common point with the river.

For convenience, scientists have made a transformation of coordinates so that the river is defined by y=0y=0. Check whether it is possible to build a reserve, and if possible, find the minimum possible radius of such a reserve.

Input

The first line contains one integer nn (1≤n≤1051≤n≤105) — the number of animals.

Each of the next nn lines contains two integers xixi, yiyi (−107≤xi,yi≤107−107≤xi,yi≤107) — the coordinates of the ii-th animal's lair. It is guaranteed that yi≠0yi≠0. No two lairs coincide.

Output

If the reserve cannot be built, print −1−1. Otherwise print the minimum radius. Your answer will be accepted if absolute or relative error does not exceed 10−610−6.

Formally, let your answer be aa, and the jury's answer be bb. Your answer is considered correct if |a−b|max(1,|b|)≤10−6|a−b|max(1,|b|)≤10−6.

Examples

input

Copy

1
0 1

output

Copy

0.5

input

Copy

output

Copy

-1

input

Copy

2
0 1
1 1

output

Copy

0.625

Note

In the first sample it is optimal to build the reserve with the radius equal to 0.50.5 and the center in (0, 0.5)(0, 0.5).

In the second sample it is impossible to build a reserve.

In the third sample it is optimal to build the reserve with the radius equal to 5858 and the center in (12, 58)(12, 58).

题意：给出n个点，找到一个圆，使得这个圆的半径最小，并且与X轴相切，并且包含n个点

思路：

1.二分找半径r，那么我们要判断r是否合法，那么可以根据已给出的r求出每个圆圆心的x的坐标区间（y坐标永远是R），那么只要所有的点的圆的x的区间至少有一个交点，那么当前的r是合法的，那么减小r的取值，如果不合法，那么说明给的R太小了，把R变大即可

#include <stdio.h>
#include<string.h>
#include<algorithm>
#include<math.h>
#define maxn 100001 +10
#define INF 0xfffffff
using namespace std;	
int n;
struct node{
	double x,y;
}a[maxn];

int check(long double r)
{	
	int i;	
	double l=-100000000000000000.0;
 	double rig=100000000000000000.0;
	for(i=0;i<n;i++)
	{
		if(a[i].y>r+r)
		{	
			return 0;
		}
		if(r*r-(a[i].y-r)*(a[i].y-r)<0)
		return 1;
	 	double left=a[i].x-sqrt(2*a[i].y*r-a[i].y*a[i].y);
	 	double right=a[i].x+sqrt(2*a[i].y*r-a[i].y*a[i].y);
		l=max(left,l);
		rig=min(right,rig);
	}
	return l<rig;
}
int main(int argc, char *argv[])
{

	while(scanf("%d",&n)!=EOF)
	{
		memset(a,0,sizeof(a));
		int i;
		int z=0;
		int f=0;
		for(i=0;i<n;i++)
		{
			scanf("%lf %lf",&a[i].x,&a[i].y);
			if(a[i].y<0)
			{
				f++;
			}
			else
			{
				z++;
			}
			if(a[i].y<0)
			{
				a[i].y=-a[i].y;
			}
		}	
		if(f>0&&z>0)
		{
			printf("-1\n");
			continue; 
		}	
		double left=0.0;
		double right=100000000000000000.0;
		double mid=(left+right)*1.0/2;
		//printf("%.10Lf\n",right);
		int cnt=0;
		while(cnt<301)
		{
			cnt++;
			//printf("%lf\n",mid);
			if(check(mid))
			{
				right=mid;
			}	
			else
			{
				left=mid;
			}
			mid=(right+left)/2.0;
		} 
		printf("%lf\n",mid);
	}	
	return 0;
}