POJ 1328 Radar Installation

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 21 2 -3 1 2 1 1 2 0 2 0 0

Sample Output

#include <bits/stdc++.h>
const double INF=0x3f3f3f3f;
using namespace std;

struct node
{
  double start, end;
};
node s[1500];

double cmp(node a, node b)//以s.end升序写的cmp；
{
  return a.end<b.end;
}
int main()
{
  int n, d, i;
  double x, y;
  int K=0;
  while(cin>>n>>d)
    {
      if(n==0||d==0)
        return 0;
      int flag=1;
      for(i=0; i<n; i++)
        {
          cin>>x>>y;
          if(y<=d)
            {
              s[i].start=x-sqrt(d*d-y*y);
              s[i].end=x+sqrt(d*d-y*y);
            }
          else
            flag=0;
        }
      cout<<"Case "<<++K<<": ";
      if(flag==0)
        {
          cout<<"-1"<<endl;
        }
      else
        {
          sort(s,s+n,cmp);//快排：s.end升序
//          for(i=0;i<n;i++)
//          printf("%.2lf  %.2lf\n",s[i].start, s[i].end);
          int sum=0;
          double temp = -INF;
          for(i=0; i<n; i++)
            {
              if(temp<s[i].start)
                {
                  sum++;
                  temp=s[i].end;
                }
            }
            cout<<sum<<endl;
        }
    }
  return 0;
}