Radar Installation

Radar Installation

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 75387		Accepted: 16872

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
越来越发现自己是只菜鸟，花了一个上午弄懂区间找点的题，刚刚开始百度真的看不懂，但是后来就慢慢明白了，像我一样的菜鸟们，加油，没有什么不可能。

#include<cstdio>
#include<cstring>
#include<cmath>
#include<cstdlib>
#include<algorithm>
using namespace std;
struct stu{
	double b,e;
}a[1100];
bool cmp(stu x,stu y)
{
	if(x.e !=y.e)
	return x.e<y.e;
	return x.b>y.b;
}
int main()
{
	int n,d,i,f,x,y,k=1,count;;
	double l,c;
	while(scanf("%d%d",&n,&d)&&(n||d))
	{
		f=0;
		for(i=0;i<n;i++)
		{
		scanf("%d%d",&x ,&y);
		if(y>d)
		{
			f=1;
			continue;
		}
		l=sqrt((d*d-y*y)*1.0);
		a[i].b=x-l;
		a[i].e=x+l;
	    }
	    printf("Case %d: ",k++);
	    if(f==1)
	    {
		printf("-1\n");
		continue;
	    }
	    sort(a,a+n,cmp);
	     c=a[0].e;
	     count=1;
	    for(i=1;i<n;i++)
	    {
	    	if(a[i].b>c)
	    	{
	    		c=a[i].e;
				++count;
			}
		}
		printf("%d\n",count);
	}
	return 0;
}