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.

输入

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

输出

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.

样例输入

3 2
1 2
-3 1
2 11 2
0 20 0

样例输出

Case 1: 2
Case 2: 1

解题思路：

此题意思给定点集S={（xi，yi）i=1.2.3...n}，求用圆心在x轴上，半径为d的圆覆盖S所需的圆的最少个数。

1.先把给出的岛的坐标（xi.yi）和半径r转化为在x轴上的区间，即当d-yi>=0时，圆心位于x轴上的区间为Ii=[ xi-sqrt(d^2-yi^2) , xi + sqrt( d^2 - yi^2 )],则转化为区间选点问题。

2.S中点（xi，yi），对应一个在x轴上的区间Ii=（li，ri），按照区间右端点ri从小到大排序，在区间集合中选择一个索引最小的区间，把选择的区间和与其相交的所有区间作为一组从T中删除，直到T为空集、

3则剩下的分组的组数即为m的最小值。。。

AC代码：

#include<cstdio>
#include<cmath>
#include<algorithm>
#define MAX 1010using namespace std;class Range
{public:float left;float right;bool operator<(const Range &r)const{if(right<r.right||(right==r.right&&left>r.left))return true;elsereturn false;}
};int main(int argc,char *argv[])
{int n,d;int x,y;int i,j;int count=1,ans;while(scanf("%d%d",&n,&d)&&(n+d)){Range r[MAX];j=0;ans=0;for(i=0;i<n;i++){Range temp;scanf("%d%d",&x,&y);if(d>=y){temp.left=x-sqrt(d*d-y*y);temp.right=x+sqrt(d*d-y*y);r[j++]=temp;}}sort(r,r+j);if(j<n)printf("Case %d: -1\n",count);else{float end=-0x0FFFFFFF;for(i=0;i<j;i++){if(end<r[i].left){ans++;end=r[i].right;}}printf("Case %d: %d\n",count,ans);}count++;}return 0;
}

解题方法跟求最大区间数极其相似。。。