POJ 1328 Radar Installation

Radar Installation

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 41153		Accepted: 9125

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 <iostream>

#include <cstdio>

#include <cstring>

#include <cmath>

#include <algorithm>

using namespace std;

int n;double d;

struct Node

{

    double left,right;

}island[1111];

bool cmp(Node a,Node b)

{

    return a.left-b.left<1e-6;

}

int main()

{

    int cas=0;double x,y;

while(scanf("%d%lf",&n,&d))

{

    memset(island,0,sizeof(island));

    if(n==0) break;

    int possible=1;

    for(int i=0;i<n;i++)

    {

        scanf("%lf%lf",&x,&y);

        if(y>d||d<0) possible=0;

        double temp=sqrt(d*d-y*y);

        island.left=x-temp;

        island.right=x+temp;

    }

    if(possible==0)

    {

        printf("Case %d: -1 ",++cas);

        continue;

    }

    sort(island,island+n,cmp);

    int ans=1;

    Node pre=island[0];

    for(int i=1;i<n;i++)

    {

        if(island.left-pre.right>1e-6)

        {

            ans++;

            pre=island;

        }

        else

        {

            if(island.right-pre.right<1e-6)

            {

                pre=island;

            }

        }

    }

     printf("Case %d: %d ",++cas,ans);

}

    return 0;

}