贪心 Radar Installation (求最少探测雷达）

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



    先算出每个岛放探测雷达的x坐标范围，b为雷达探测半径，岛的坐标为（x，y），雷达的坐标范围是（x-sqrt(d*d-y*y),x+sqrt(d*d-y*y)），定义sum=1，i从1开始，每个坐标范围左右边界与上一个坐标范围的右边界比较，具体看代码。

#include<cstdio>
#include<algorithm>
#include<cmath>
using namespace std;
struct stu
{
    double l,r;
} st[1500];
bool cmp(stu a,stu b)
{
    return a.l<b.l;
}
int main()
{
    int a,s,k=0,sum,i;
    double b,x,y,sky;
    while(scanf("%d %lf",&a,&b) &&!(a==0&&b==0))
    {
        s=0;
        k++;
        for(i = 0 ; i < a ; i++)
        {
            scanf("%lf %lf",&x,&y);
            if(b < 0 || y > b)
            {
                s=-1;
            }
            st[i].l=x-sqrt(b*b-y*y);
            st[i].r=x+sqrt(b*b-y*y);
        }
        if(s == -1) printf("Case %d: -1\n",k);
        else
        {
            sort(st,st+a,cmp);
            sum=1;
            sky=st[0].r;
            for(i = 1 ; i < a ; i++)
            {
                if(st[i].r <= sky)
                {
                    sky=st[i].r;
                }
                else
                {
                    if(st[i].l > sky)
                    {
                        sky=st[i].r;
                        sum++;
                    }
                }
            }
            printf("Case %d: %d\n",k,sum);
        }
        
    }
}