POJ 1328 Radar Installation

Radar Installation

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 38000		Accepted: 8443

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;

const int N=1010;

struct node{
    double l,r;
}seg[N];

int cmp(node a,node b){
    return a.l<b.l;
}

int main(){

    //freopen("input.txt","r",stdin);

    int n,d,cases=0;
    while(~scanf("%d%d",&n,&d)){
        if(n==0 && d==0)
            break;
        int x,y;
        double tmp;
        int flag=0;
        for(int i=0;i<n;i++){
            scanf("%d%d",&x,&y);
            tmp=sqrt((double)(d*d)-y*y);
            seg[i].l=x-tmp;
            seg[i].r=x+tmp;
            if(y>d)
                flag=1;
        }
        sort(seg,seg+n,cmp);
        printf("Case %d: ",++cases);
        if(flag){
            printf("-1\n");
            continue;
        }
        int ans=1;
        node line=seg[0];
        for(int i=1;i<n;i++){
            if(line.r>=seg[i].r)
                line=seg[i];
            else if(line.r<seg[i].l){
                ans++;
                line=seg[i];
            }

        }
        printf("%d\n",ans);
    }
    return 0;
}