[POJ] 1328 - Radar Installation

Radar Installation

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 44705		Accepted: 9913

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

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#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<math.h>
using namespace std;
struct P{
    double x,y;
}s[1005],d[1005];
int cmp(P a,P b){
    if(a.x!=b.x)
    return a.x<b.x;
    return a.y>b.y;
}
int main(){
    int n,i,j,ans,flag,C=0;
    double p,m;
    while(scanf("%d%lf",&n,&m)!=EOF && (n!=0 || m!=0)){
        flag=0;
        for(i=0;i<n;i++){
            scanf("%lf%lf",&s[i].x,&s[i].y);
            d[i].x=s[i].x-sqrt(m*m-s[i].y*s[i].y);
            d[i].y=s[i].x+sqrt(m*m-s[i].y*s[i].y);
            if(s[i].y<0 || s[i].y>m){
                flag=1;
            }
        }
        if(flag){
            printf("Case %d: -1\n",++C);
            continue;
        }
        ans=0;
        sort(d,d+n,cmp);
        flag=0;
        for(i=0;i<n;i++){
            if(i && d[i].y<p)
                p=d[i].y;
            if(i==0 || d[i].x>p)
                p=d[i].y,ans++;

        }
        printf("Case %d: %d\n",++C,ans);
    }
}