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 12 11 20 20 0

Sample Output

Case 1: 2Case 2: 1

题目大意（转）：是为了求出能够覆盖所有岛屿的最小雷达数目，每个小岛对应x轴上的一个区间，在这个区间内的任何一个点放置雷达（在x轴上），则可以覆盖该小岛，区间范围的计算用[x-sqrt(d*d-y*y),x+sqrt(d*d-y*y)]。

这样，问题即转化为已知一定数量的区间，求最小数量的点，使得每个区间内斗至少存在一个点。每次迭代对于第一个区间， 选择最右边一个点， 因为它可以让较多区间得到满足，。如果不选择第一个区间最右一个点（而选择前面的点）， 那么把它换成（前面区间）最右的点之后， 以前得到满足的区间， 现在仍然得到满足。

 

1 #include 
     
       2 #include
      
        3 #include
       
         4 #include
        
          5 using namespace std; 6  7 #define maxn 1001 8  9 struct node10 {11     double left;12     double right;13 }island[maxn];14 15 bool cmp(node a,node b)16 {17     return a.left
         
          d||d<0)33             {34                 flag=1;35             }36             island[i].left=x-sqrt(d*d-y*y);//先算出每个岛的雷达覆盖区间37             island[i].right=x+sqrt(d*d-y*y);//先算出每个岛的雷达覆盖区间38         }39         if(flag)40         {41             printf("Case %d: -1\n",k++);42             continue;43         }44         sort(island,island+n,cmp);//按左端点排序45         temp=island[0].right;//排序完成后，第一个雷达建立在第一个区间的右端点46         count=1;47         for(i=1;i
          
           temp)//判断每个区间的左端点，如果在最新建立的雷达右边54 {55 count++;//那么肯定需要一个新雷达56 temp=island[i].right;//而且也建在该区间右端57 }58 }59 printf("Case %d: %d\n",k++,count);60 }61 return 0;62 }