Radar Installation

Radar Installation

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 77204		Accepted: 17294

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

Source

Beijing 2002

#include<iostream>
#include<cmath>
#include<string.h>
#include<algorithm>
using namespace std;
const int Max = 1001;
struct node
{
	double l;
	double r;
}s[Max];
bool cmp(node a,node b)
{
	if(a.r < b.r)
		return true;
	else if(a.r == b.r)
	{
		if(a.l > b.l)
			return true;
		else
			return false;
	}
	return false;
}
int main()
{
	int n,d;
	double x,y;
	int t = 1;
	while(cin>>n>>d && (n||d))
	{
		bool f = 1;
		for(int i = 1;i <= n;i++)
		{
			cin>>x>>y;
			double temp = d*d - y*y;
			if(temp < 0 || d < 0)
				f = 0;
			else if(f)
			{
				s[i].l = x - sqrt(1.0*d*d - y*y*1.0);
				s[i].r = x + sqrt(1.0*d*d - y*y*1.0);
			}
		}
		if(!f)
		{
			cout<<"Case "<<t++<<": "<<-1<<endl;
			continue;
		}
		sort(s+1,s+1+n,cmp);
		int sum = 1;
		double temp = s[1].r;
		for(int i = 2;i <= n;i++)
		{
			if(temp < s[i].l)
			{
				sum++;
				temp = s[i].r;
			}
		}
		cout<<"Case "<<t++<<": "<<sum<<endl;
	}
	return 0;
}