Radar Installation POJ - 1328

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 21 2-3 12 11 20 20 0

Sample Output

Case 1: 2Case 2: 1

#include <iostream>#include <cstdio>#include <cmath>#define MAX 1010struct node{    double x,y,l,r;}a[MAX],temp;int main(){    int n;    double r;    int num=0;    while(~scanf("%d %lf",&n,&r))    {        num++;        int flag=1;        if(n==0&&r==0)            break;        for(int i=1;i<=n;i++)        {            scanf("%lf %lf",&a[i].x,&a[i].y);            a[i].l=a[i].x-sqrt(r*r-a[i].y*a[i].y);            a[i].r=a[i].x+sqrt(r*r-a[i].y*a[i].y);            if(a[i].y>r)            {                flag=0;//此次尽量不标break，没有输入完            }        }        if(flag==0)        {            printf("Case %d: -1\n",num);            continue;        }        for(int i=1;i<=n;i++)        {            for(int j=1;j<=n-1;j++)            {                if(a[j].x>a[j+1].x)                {                    temp=a[j];                    a[j]=a[j+1];                    a[j+1]=temp;                }            }        }             int ans=1,pos=1;        for(int i=2;i<=n;i++)        {            if(a[pos].r<a[i].l)            {                ans++;                pos=i;            }            else if(a[i].r<a[pos].r)//当范围在其最右端的左边切记，考虑全面            {                pos=i;            }        }        printf("Case %d: %d\n",num,ans);    }    return 0;}