hdoj Radar Installation

Problem 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
#include<stdio.h>
#include<math.h>
#include<algorithm>
using namespace std;
struct node
{
	 float l,r,x,y;
}d[1100];
int cmp(node x,node y)
{
	return x.l<y.l;/*按照区域的最左端排序*/
}
int main()
{
	int sum,t,m,n,flog=1,Case=1,i,num,j;
	while(scanf("%d%d",&m,&n),m||n)
	{
		for(i=0;i<m;i++)
		{
			scanf("%f%f",&d[i].x,&d[i].y);
			if(d[i].y>n) 
			{
			flog=0;break;
			}
			else 
			{
				d[i].l=d[i].x-sqrt(n*n-d[i].y*d[i].y);/*求出雷达的扫描区域*/
				d[i].r=d[i].x+sqrt(n*n-d[i].y*d[i].y);
			}
		}
		printf("Case %d: ",Case);
		if(flog==0) printf("-1
");
		else 
		{
			int sign=0;num=0;
			sort(d,d+m,cmp);
			num++;sign=d[0].r;/*这个很重要，sign始终标记雷达所能扫描的最右端*/
			for(i=1;i<m;i++)/*至少设置一个雷达，如果当前判断的区域达不到雷达最右端，加设一个雷达*/
			{
				if(d[i].r<sign)
				sign=d[i].r;
				else if(d[i].l>sign)
				{
					num++;sign=d[i].r;
				}
			}
			printf("%d
",num);
		}
	}
	return 0;
}