Radar Installation

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

海岛和雷达问题，看看怎样用最少的雷达覆盖所有海岛；
输入 3 2 是海岛数 和雷达覆盖半径；
接下来是 ： 海岛坐标
我感觉这题好奇葩，测试数据都过了，就是提交不上，说明你还是有地方用了整形（%d）,看看能换的全部用double 试试；
下面是我同学的代码……我的是数据都过 ……就是WA

#include <stdio.h>
#include<string.h>
#include<stdlib.h>
#include<math.h>
using namespace std;
struct node
{
   double a,b;
}area[1005];
int cmp(const void *a,const void *b)
{
    return (*(node *)a).b>(*(node *)b).b?1:-1;
}
int main()
{
    int n,i,sum,f,h=1;
    double x,y,d,e;
    while(~scanf("%d%lf",&n,&d)&&(n!=0||d!=0))
    {
         f=0;   sum=1;
        for(i=0; i<n; i++)
        {
            scanf("%lf%lf",&x,&y);
            area[i].a=x-sqrt(d*d-y*y);//这里是坐标转区间
            area[i].b=x+sqrt(d*d-y*y);
            if(y<0)
            y=-y;
            if(y>d)
            f=1;
        }

        if(f)
        printf("Case %d: -1
",h);

        else
        {
            qsort(area,n,sizeof(area[0]),cmp);
            e=area[0].b;
            for(i=1; i<n; i++)
            {
                if(area[i].a>e)//找左边界小于该右边界的
                {
                    sum++;
                    e=area[i].b;
                }
            }
            printf("Case %d: %d
",h,sum);
        }
        h++;
    }
    return 0;
}