E - Radar Installation

Assume the coasting is an infinite straight line. Land is in one side of coasting, sea in the other. Each small island is a point locating in the sea side. And any radar installation, locating on the coasting, can only cover d distance, so an island in the sea can be covered by a radius installation, if the distance between them is at most d.

We use Cartesian coordinate system, defining the coasting is the x-axis. The sea side is above x-axis, and the land side below. Given the position of each island in the sea, and given the distance of the coverage of the radar installation, your task is to write a program to find the minimal number of radar installations to cover all the islands. Note that the position of an island is represented by its x-y coordinates.

Figure A Sample Input of Radar Installations

Input

The input consists of several test cases. The first line of each case contains two integers n (1<=n<=1000) and d, where n is the number of islands in the sea and d is the distance of coverage of the radar installation. This is followed by n lines each containing two integers representing the coordinate of the position of each island. Then a blank line follows to separate the cases.

The input is terminated by a line containing pair of zeros

Output

For each test case output one line consisting of the test case number followed by the minimal number of radar installations needed. "-1" installation means no solution for that case.

Sample Input

3 2

1 2

-3 1

2 1

1 2

0 2

0 0

Sample Output

Case 1: 2

Case 2: 1

想要按照最少的雷达站，先把x排序，如何判断下一个的最左边是不是在之前的雷达的最右边的右边，是就要重新按一个雷达站；

要注意，如果下一个的最右边在之前雷达的左边，说明安装的位置要在这里的最右边；

#include<iostream>
#include<stdio.h>
#include<stdlib.h>
#include<cmath>
#include<string.h>
#include<algorithm>
#define sf scanf
#define pf printf
#define cl clear()
#define pb push_back
#define mm(a,b) memset((a),(b),sizeof(a))
#include<vector>
const double pi=acos(-1.0);
typedef __int64 ll;
typedef long double ld;
const ll mod=1e9+7;
using namespace std;
int n,d,bits=1;
double wzl(int y)
{
	double x=d*d-y*y;
	return sqrt(x);
}
int a[1005],b[1005];
int main()
{

	while(1)
	{
		mm(a,0);
		mm(b,0);
		sf("%d%d",&n,&d);
		if(n==0)
		return 0;
		int temp=0;
		for(int i=0;i<n;i++)
		{
			sf("%d%d",&a[i] ,&b[i] );
			if(b[i]>d)
			temp=1;
		}
		if(temp)
		{
			pf("Case %d: -1\n",bits++);
			continue;
		}
		for(int i=0;i<n-1;i++)
		for(int j=0;j<n-i-1;j++)
		{
			if(a[j]>a[j+1])
			{
				int w=a[j];
				a[j]=a[j+1];
				a[j+1]=w;
				w=b[j];
				b[j]=b[j+1];
				b[j+1]=w;
			}
		}
		int sum=1;
		double last=a[0]+wzl(b[0]),left,right;
		for(int i=1;i<n;i++)
		{
			left=a[i]-wzl(b[i]);
			right=a[i]+wzl(b[i]);
			if(left>last)
			{
				sum++;
				last=right;
			}else if(right<last)
			{
				last=right;
			}
		}
		pf("Case %d: %d\n",bits++,sum);
	}
}