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.

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std;
const int M = 1001;
structxy
{
    double l;
    double r;
}p[M];

double min(double a,double b)
   {
      return a < b ? a :b;
   }
int d,n;
bool cmp(structxy a,structxy b )
{
    return a.l < b.l;
}
int main()
{
    double a,b;
    int icase = 0;
    while(cin >> n >> d && n && d)
    {
        memset(p,0,sizeof(0));
        icase++;
        int sum = 1;
        bool flag = true;
        for(int i = 1; i <= n;i++)
        {
           cin >>a>>b;
           if(b > d)
           {
               flag = true;
           }
           else
           {
              p[i].l = a - sqrt(d*d - b*b);
              p[i].r = a + sqrt(d*d - b*b);
           }

        }
        if(!flag)
        {
            cout<<"Case "<<icase<<": -1"<<endl;
            continue;
        }
        else
        {
            sort(p+1,p+n+1,cmp);
            double s = p[1].r;
            for(int i = 2; i <= n; i++)
            {
                if(p[i].l > s)
                {
                    sum++;
                    s = p[i].r;
                }
                else
                {
                    s = min(s,p[i].r);
                }
            }
        }
      cout<<"Case "<<icase<<": "<<sum<<endl;
      cout<<endl;
    }
    return 0;
}