Problem Description

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

#include
      
       #include
       
        #include
        
         using namespace std;struct node{	 float l,r,x,y;}d[1100];int cmp(node x,node y){	return x.l
         
          n) 			{			flog=0;break;			}			else 			{				d[i].l=d[i].x-sqrt(n*n-d[i].y*d[i].y);/*求出雷达的扫描区域*/				d[i].r=d[i].x+sqrt(n*n-d[i].y*d[i].y);			}		}		printf("Case %d: ",Case);		if(flog==0) printf("-1\n");		else 		{			int sign=0;num=0;			sort(d,d+m,cmp);			num++;sign=d[0].r;/*这个很重要，sign始终标记雷达所能扫描的最右端*/			for(i=1;i
          
           sign)				{					num++;sign=d[i].r;				}			}			printf("%d\n",num);		}	}	return 0;}