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 InstallationsInput

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

分析：对小岛与海岸线的交点按从左到右的次序排序，排序完成后，第一个雷达建立在第一个区间的右端，然后判断每个区间的左端点，如果在最新建立的雷达右面，那么肯定需要一个雷达，而且也建在区间右端。如果左端点在雷达左面，这个时候要考虑区间的右端在雷达的左面还是右面，如果是右面，那雷达位置就不变，如果在左面，那现在的雷达是覆盖不了的，所以要把雷达放在该区间的右端点！因为这样同时还能覆盖原来的岛，还能覆盖现在的岛！

code：

#include
      
        #include
       
         #include
        
          struct di {
    double l; double r; }q[1001]; int cmp(const void*p1,const void*p2) {
    struct di *c=(struct di*)p1; struct di *d=(struct di*)p2; return c->l>=d->l?1:-1; } int main() {
    int tot,n,i,c=1,flag; double r,x,y,be; while(scanf("%d%lf",&n,&r),n||r)     {
            flag=0; for(i=0;i
         
          r)                 flag=1;             q[i].l=x-sqrt(pow(r,2)-pow(y,2));             q[i].r=x+sqrt(pow(r,2)-pow(y,2));         }         qsort(q,n,sizeof(q[0]),cmp);         tot=1;         be=q[0].r; for(i=1;i
          
           be)             {
                    tot++;                 be=q[i].r;             } else if(q[i].r