Radar Installation

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<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
using namespace std;
struct node
{
	double zuo,you;
}lei[2001];
bool cmp(node a,node b)
{
	return a.zuo  <b.zuo  ;
}
int main()
{
     int d,n,cut=0;
	while(scanf("%d%d",&n,&d)!=EOF)
	{
		if(n==0 && d==0)
            break;
        int x,y;
        double tmp;
        int flag=0;
        for(int i=0;i<n;i++){
            scanf("%d%d",&x,&y);
            tmp=sqrt((double)(d*d)-y*y);
            lei[i].zuo=x-tmp;
            lei[i].you=x+tmp;//求出每个岛能被扫到的雷达所在区间  
            if(y>d)
                flag=1;
        }	
		 	if(flag)
		 	{
		 			printf("Case %d: -1
",++cut);
		 			continue;
			 }
		sort(lei,lei+n,cmp);
		int sum=1;
		node  qian= lei[0];
		for(int i=1;i<n;i++)
		{
				 if( qian.you  < lei[i].zuo  )
			         {    
					     sum++;
			         	qian=lei[i] ;
		 	         }
			         else
			         {
			         	    if(lei[i].you<=qian.you )
			               {
			          	      qian=lei[i] ;
						    }
					 }
			  }
				printf("Case %d: %d
",++cut,sum);
	 } 
	return 0;
}