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 题意：将一条海岸钱看为X轴，X轴的上方为大海，海上有许多岛屿，给出岛屿的位置与雷达的覆盖半径，要求在海岸线上建雷达，
在雷达能够覆盖所有岛的基础上，求最少需要多少雷达。
#include <iostream>
#include <cstdio>
#include <cmath>
#include <algorithm>
#include <cstring>
#include <cstdlib>
using namespace std;
#define MAX 1005
struct sea
{
double left;
double right;
} a[];
bool operator < (sea A,sea B)
{
return A.left<B.left;
}
int main()
{
int n,k=;
double d;
while(cin>>n>>d&&(n||d))
{
bool flag=false;
for(int i=; i<n; i++)
{
double x,y;
cin>>x>>y;
if(fabs(y)>d)
flag=true;
else
{ //计算区间
a[i].left=x*1.0-sqrt(d*d-y*y);
a[i].right=x*1.0+sqrt(d*d-y*y);
}
}
printf("Case %d: ",k++);
if(flag)
printf("-1\n");
else
{
int ans=; //雷达初始化
sort(a,a+n); // 排序
double s=a[].right;
for(int i=; i<n; i++)
{
if(a[i].left>s)
{
ans++; //雷达加一
s=a[i].right; // 更新右端点
}
else if(a[i].right<s)
s=a[i].right;
}
printf("%d\n",ans);
}
}
return ;
}