ACM Radar Installation(挑战程序设计竞赛)

Radar Installation

Time Limit: 1000ms

Memory Limit: 10000KB

This problem will be judged on PKU. Original ID: 1328
64-bit integer IO format: %lld      Java class name: Main

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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 
#include 
#include 
#include 

using namespace std;

#define INF 99999999
#define MAX_N 1000

struct Point
{
	double l,r;
} pt[MAX_N];

int n,d;
bool bNo;

bool cmp(const Point &a,const Point &b)
{
	return a.ld)
			{
				bNo=true;
				continue;
			}
			double delta=sqrt(d*d-y*y);
			pt[i].l=x-delta;
			pt[i].r=x+delta;
		}

		if(!bNo)
		{
			sort(pt,pt+n,cmp);

			double maxr;

			maxr=-INF;
			int cnt=0;
			for(int i=0;ipt[i].r)
				{
					maxr=pt[i].r;
				}
			}
			printf("Case %d: %d\n",T++,cnt);
		}
		else
			printf("Case %d: -1\n",T++);

		scanf("%d%d",&n,&d);
	}

	return 0;
}