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<stdio.h>

#include<math.h>

#include<algorithm>

using namespace std;

struct node

{

	 float l,r,x,y;

}d[1100];

int cmp(node x,node y)

{

	return x.l<y.l;/*按照区域的最左端排序*/

}

int main()

{

	int sum,t,m,n,flog=1,Case=1,i,num,j;

	while(scanf("%d%d",&m,&n),m||n)

	{

		for(i=0;i<m;i++)

		{

			scanf("%f%f",&d[i].x,&d[i].y);

			if(d[i].y>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<m;i++)/*至少设置一个雷达，如果当前判断的区域达不到雷达最右端，加设一个雷达*/

			{

				if(d[i].r<sign)

				sign=d[i].r;

				else if(d[i].l>sign)

				{

					num++;sign=d[i].r;

				}

			}

			printf("%d\n",num);

		}

	}

	return 0;

}