Q - Radar Installation 【区间选点】

Q - Radar Installation

POJ - 1328 

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

思路： 转换看图的方式，。。。

区间选点问题，  多个区间，问至少要几个点才可以 在每个区间里都有点存在

代码

#include
#include
#include
typedef struct
{
	double left,right;
}data;
data a[1000+10];
bool cmp(data a,data b)
{
	if(a.right!=b.right) return a.rightb.left;
 } 
 using namespace std;
 int main()
 {
 	int n,d;int p=1;
 	while(scanf("%d%d",&n,&d)&&(n!=0||d!=0))
 	{
 		int l=1;
 		for(int i=0;i