POJ 1328 Radar Installation

Radar Installation

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 49398		Accepted: 11037

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 <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <math.h>
#include <ctype.h>
#include<algorithm>

using namespace std;

struct node
{
    double x,y;
} a[1001];

bool cmp(node x,node y)
{
    return x.x<y.x;
}

int main()
{
    int n,d,i,j,x,y,l=0;
    while(scanf("%d%d",&n,&d)!=EOF&&(n!=0&&d!=0))
    {
        l++;
        int flag=0;
        for(i=0; i<n; i++)
        {
            scanf("%d%d",&x,&y);
            if(y>d)
                flag=1;
            a[i].x=x-sqrt(d*d-y*y);
            a[i].y=x+sqrt(d*d-y*y);
        }
        if(flag)
        {
            printf("Case %d: -1\n",l);
            continue;
        }
        sort(a,a+n,cmp);
        int ans=1;
        double border=a[0].y;
        for(i=0; i<n; i++)
        {
            if(a[i].x>border)
            {
                ans++;
                border=a[i].y;
            }
            if(a[i].y<border)
                border=a[i].y;
        }
        printf("Case %d: %d\n",l,ans);
    }
    return 0;
}