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

Sample Output

分析：

简单的贪心，先按x轴排序，记录圆心位置范围，如果一个点的圆心范围和前面一个圆心范围有交集，就把前一个圆心范围更新为他们的交集

不然答案+1，将当前圆心范围记录下来，最后输出ans

#include 
     
      #include 
      
       #include 
       
        #include 
        
         using namespace std;struct node {	int x, y;};node g[1111];double posr[1111], posl[1111];int n, d, ans;int cmp (node a, node b) {	if (a.x == b.x) return a.y > b.y;	return a.x < b.x;}double fx (node x) {	double k = sqrt (1.*d * d - x.y * x.y);	return k;}int main() {	for (int t = 1; cin >> n >> d; t++) {		if (n == 0 && d == 0) break;		ans = 0;		for (int i = 1; i <= n; i++) {			cin >> g[i].x >> g[i].y;			if (g[i].y > d) ans = -1;		}		if (ans == 0) {			sort (g + 1, g + 1 + n, cmp);			if (n >= 1) {				posl[++ans] = g[1].x - fx (g[1]);				posr[ans] = g[1].x + fx (g[1]);			}			for (int i = 2; i <= n; i++) {				if (g[i].x == g[i - 1].x) continue;				if (g[i].x - fx (g[i]) > posr[ans]) {					posl[++ans] = g[i].x - fx (g[i]), posr[ans] = g[i].x + fx (g[i]);					continue;				}				posl[ans] = max (posl[ans], g[i].x - fx (g[i]) ), posr[ans] = min (posr[ans], g[i].x + fx (g[i]) );			}		}		printf ("Case %d: %d\n", t, ans);	}	return 0;}