POJ-1328-Radar Installation-2013-12-07 01:49:28

Radar Installation

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 45922		Accepted: 10252

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

Source

Beijing 2002

# include<stdio.h>
# include<string.h>
# include<math.h>
# include<iostream>
# include<algorithm>
using namespace std;

typedef struct Point
{
	double left;
	double right;
}p;

int cmp(p x,p y)
{
	return x.left < y.left;
}

int main()
{
	int i,n,k,r,x,y,flag,sum;
	double t;
	p point[1010];
	k = 0;
	while(~scanf("%d%d",&n,&r) && n!=0 || r!=0)
	{
		flag = 1;
		for(i=0;i<n;i++)
		{
			scanf("%d%d",&x,&y);
			if(abs(y)>r)
				flag = 0;

			point[i].left = x*1.0 - sqrt(r*r*1.0 - y*y*1.0);
			point[i].right = x*1.0 + sqrt(r*r*1.0 - y*y*1.0);
		}
		
		sort(point,point+n,cmp);
		
		sum = 1;
		t = point[0].right;
		for(i=1;i<n;i++)
		{
			if(point[i].right<t)
			{
				t = point[i].right;
			}
			else if(point[i].left>t)
			{
				t = point[i].right;
				sum++;
			}
		}

		printf("Case %d: ",++k);

		if(flag==0)
			printf("-1\n");
		else
			printf("%d\n",sum);

	}

	return 0;
}