Radar Installation

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 56702		Accepted: 12792

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 21 2-3 12 11 20 20 0

Sample Output

Case 1: 2Case 2: 1 思路就是，排序之后，按右端点放雷达就好

#include
     
      #include
      
       #include
       
        #include
        
         #include
         
          using namespace std;const int N=1010;struct node{    double l,r;}seg[N];int cmp(node a,node b){    return a.l
          
           d) flag=1; } sort(seg,seg+n,cmp); printf("Case %d: ",++cases); if(flag){ printf("-1\n"); continue; } int ans=1; node line=seg[0]; for(int i=1;i
           
            =seg[i].r) line=seg[i]; } printf("%d\n",ans); } return 0;}