Radar Installation

Time Limit: 1000MS	Memory Limit: 10000K
Total Submissions: 80210	Accepted: 17922

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
解題思路：該題題意是為了求出能夠覆蓋所有島嶼的最小雷達數目，每個小島對應x軸上的一個區間，在這個區間內的任何一個點放置雷達，則可以覆蓋該小島，區間範圍的計
算用[x-sqrt(d*d-y*y),x+sqrt(d*d-y*y)];這樣，問題即轉化為已知一定數量的區間，求最小數量的點，使得每個區間內鬥至少存在一個點。每次迭代對於第一個區間，選
擇最右邊一個點， 因為它可以讓較多區間得到滿足， 如果不選擇第一個區間最右一個點（選擇前面的點）， 那麼把它換成最右的點之後，以前得到滿足的區間， 現在仍然
得到滿足， 所以第一個區間的最右一個點為貪婪選擇， 選擇該點之後， 將得到滿足的區間刪掉，進行下一步迭代， 直到結束。 

#include 
#include 
#include 
#include 
using namespace std;
struct node{
	double left;
	double right;
}point[1005];
int cmp(node a,node b){
	return a.leftd){
				flag=1;
			}
		}
		if(flag){
			printf("Case %d: -1\n",cas++);
		}
		else{
			sort(point,point+n,cmp);
			temp=point[0].right;
			for(int i=1;itemp){
					sum++;
					temp=point[i].right;
				}
			}
			printf("Case %d: %d\n",cas++,sum);
		}
	}
}