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：给出坐标点的总数n和雷达辐射半径R，再给你n个坐标，问在X轴上最少需要多少个雷达可以覆盖所有坐标点。思路：待更新。。。。
#include<stdio.h>
#include<stdlib.h>
#include<math.h>
#include<algorithm>
using namespace std;
#define N 1100
struct node {
double x1,x2;
}c[N];
double cmp(struct node a,struct node b)
{
if(a.x1 > b.x1 )
return a.x1 < b.x1 ;
}
int main()
{
int i,j,n,R,ans,t=0;
double x,y,coords;
while(scanf("%d%d",&n,&R),n!=0&&R!=0)//不能用(n+R)作为循环条件，因为3 -3这种情况不满足
{
ans = 1;
if(R <= 0)//先判断半径是否满足条件
ans = -1;
for(i = 1; i <= n; i ++)
{
scanf("%lf%lf",&x,&y);//坐标是实数
if(fabs(y) > R)//纵坐标大于半径时 不满足条件
{
ans = -1;
}
else
{
coords = sqrt(R*R-y*y);
c[i].x1 = x - coords;//计算与x轴相交的左右端点
c[i].x2 = x + coords;
}
}
if( ans == -1)
{
printf("Case %d: %d\n",++t,ans);
continue;
}
sort(c+1,c+1+n,cmp);//结构体排序
coords = c[1].x2 ;//初始化为最左坐标的右端点
for(i = 2; i <= n; i ++)
{
if(c[i].x1 > coords)//如果下一个坐标的左端点大于上一个坐标右端点 说明两者没有公共区间
{
ans ++;
coords = c[i].x2 ;
}
else if(c[i].x2 < coords)//如果下一个坐标的右端点小于上一个坐标的右端点，说明存在公共区间
coords = c[i].x2 ;
}
printf("Case %d: %d\n",++t,ans);
}
return 0;
}