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

Output

Sample Input

3 2
1 2
-3 1
2 1

1 2
0 2

0 0

Sample Output

Case 1: 2
Case 2: 1

将点转化为区间，以区间右端点进行排序。

AC代码如下：

#include<iostream>
#include<cstdio>
#include<cmath>
#include<algorithm>
using namespace std;

struct H{
    double l,r;
}a[1005];

bool cmp(H a,H b)
{
    return a.r<b.r;
}

int main()
{
    int n,r,bj;
    double h,z;
    int i,j,cas=0;
    while(cin>>n>>r,n!=0&&r!=0)
    {
        bj=0;
        cas++;
    for(i=0;i<n;i++)
    {
       cin>>h>>z;
       if(z>r)
       {
           bj=1;
       }
       a[i].l = h - sqrt(r*r-z*z);
       a[i].r = h + sqrt(r*r-z*z);
    }
    if(bj)
    {cout<<"Case "<<cas<<": "<<"-1"<<endl;continue;}
    sort(a,a+n,cmp);
    int ans=1;
    double L,R;
    L=a[0].l;
    R=a[0].r;
    for(i=1;i<n;i++)
    {
        if(a[i].l>R)//想清楚这一步，就不难了
        {
            ans++;
            R=a[i].r;
        }
    }
    cout<<"Case "<<cas<<": "<<ans<<endl;
    }
    return 0;
}