poj1328radar installation 【贪心】

Radar Installation

Time Limit : 2000/1000ms (Java/Other) Memory Limit : 20000/10000K (Java/Other)
Total Submission(s) : 93 Accepted Submission(s) : 46
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 【解题思路】：以小岛为圆心，以d为半径作圆，圆与x轴的两个交点之间的部分即为雷达可以放置的位置，左交点记为l，右交点记为r，比较区域重合的部分，与nyoj14【会场安排问题】、hdu2037【今年暑假不AC】是同种类型。 
  【代码】[code]#include<stdio.h>
#include<math.h>
#include<algorithm>
using namespace std;
#define maxn 1100
struct radar
{
	double l,r;
}a[maxn];
bool cmp(radar x,radar y)
{
	if(x.r==y.r)
	  return x.l<y.l;
	else 
	  return x.r<y.r;
}
int main()
{
	int n,d,i,x,y,flag,count;
	int k=1;
	double temp;
	while(scanf("%d%d",&n,&d)&&(n||d))
	{
		flag=0;
		for(i=0;i<n;i++)
		{
			scanf("%d%d",&x,&y);
			if(y>d)
		    {
			    flag=1;
			    continue;
		    }
		    temp=sqrt(d*d*1.0-y*y);
		    a[i].l=x-temp;
		    a[i].r=x+temp;
        }
		
	    printf("Case %d: ",k++);
	    if(flag)
	    {
	    	printf("-1\n");
	    	continue;
		}
		sort(a,a+n,cmp);
		count=1;
		temp=a[0].r;
		for(i=1;i<n;i++)
		{
			if(a[i].l>temp)
			{
				temp=a[i].r;
				count++;
			}
			  
		}
		printf("%d\n",count);
	}
	return 0;
}

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