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

大意：
有N个在X轴上方坐标已知的岛屿，然后有半径为d的雷达，只能放置在x轴上，问最少需要放置几个雷达才能覆盖所有的岛屿。如果无法覆盖输出-1。(Uva原题好像是这样的，不过听说POJ会输出在x轴下方的点?)

思路:
可以通过勾股定理算出点到圆心的距离在x轴上的投影，然后可以算出最左端和最右端的能覆盖该岛屿的雷达，确定每个岛屿能被覆盖的雷达放置区间(雷达被放在这个区间内可以覆盖该岛屿)。
之后对左端点进行排序，当前一个区间的末尾不能覆盖后一个区间的前端时，说明得多放一个雷达。
值得注意的是这里有个坑，当大区间包含一个小区间的时候，雷达放置在大区间的末尾可能覆盖不了小区间的末尾，产生错误解，所以要注意加个检测，当发现更左边的末端时，更新末端为它。

代码

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cmath>
using namespace std;
struct point{
    double x, y, s, e;
};
bool cmp(point a,point b){
    if(a.s==b.s){
        return a.e < b.e;
    }
    else return a.s < b.s;
}
int main(){
    int c=1;
    while(1){
        int n, i, ans = 0, f = 0;
        double r;
        cin >> n >> r;
        if (r == 0 && n == 0)
        {
            break;
    }
    point a[n];
    for (i = 0; i < n;i++){
        scanf("%lf%lf", &a[i].x,&a[i].y);
        if(a[i].y>r||a[i].y<0){
            f = 1;
        }
        a[i].s = a[i].x - sqrt(r * r - a[i].y * a[i].y);
        a[i].e = a[i].x + sqrt(r * r - a[i].y * a[i].y);
    }
    if(f){
        cout << "Case " << c << ": " << -1 << endl;
        c++;
        continue;
    }
    sort(a, a + n, cmp);
    i = 0;
    while(i<n){
        double s = a[i].e;
        while (a[i].s<=s&&i<n){
            if(a[i].e<s){
                s = a[i].e;
            }
            i++;
        }
        ans++;
    }
    cout << "Case " << c << ": " << ans << endl;
    c++;
    }
    return 0;
}