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.

Input 

The input consists of several test cases. The first line of each case contains two integers n (1n1000) 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

題目解法：
明顯地會發現，考慮一個最左邊的點(小島)，這個點一定在圓上，會盡可能覆蓋右方的點。
然後把可覆蓋的點都去掉，重覆這個步驟。

如果轉換成區間操作的時候，要記得是對點做排序，而非對區間做排序。
區間可以當作這個點，偵測器可以放置的可行解，因此要找有一個交集的最大量集合。

會有大的區間覆蓋小的區間，也就是 [()] 的樣子，那麼必須將右範圍縮小。

#include <stdio.h>
#include <math.h>
#include <algorithm>
using namespace std;

int main() {
    int n, cases = 0;
    int i;
    double x, y, d;
    pair<double, double> pt[1005];
    while(scanf("%d %lf", &n, &d) == 2) {
        if(n == 0)  break;
        int flag = 0;
        for(i = 0; i < n; i++) {
            scanf("%lf %lf", &x, &y);
            pt[i] = make_pair(x, y);
            if(y > d)
                flag = 1;

        }
        /*

12 21
27 1
16 19
12 6
-12 17
-2 16
-49 4
-13 12
46 12
48 4
-46 6
26 14
-22 10
        */
        if(flag) {
            printf("Case %d: %d\n", ++cases, -1);
            continue;
        }
        sort(pt, pt+n);
        for(i = 0; i < n; i++) {
            x = pt[i].first, y = pt[i].second;
            double td = sqrt(d*d - y*y) + 1e-8;
            pt[i] = make_pair(x-td, x+td);
        }
        int ret = 0;
        double r = -(1e+60);
        i = 0;
        while(i < n) {
            while(i < n && pt[i].first <= r) {
                r = min(r, pt[i].second);
                i++;
            }
            r = pt[i].second;
            ret++;
        }
        printf("Case %d: %d\n", ++cases, ret-1);
    }
    return 0;
}