poj 1328 Radar Installation

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

注意变量的初始化位置，最好在使用前初始化，如果出现错误，很难查找。
注意cmp函数的使用

#include<iostream>
#include<cmath>
using namespace std;
 
 typedef struct
 {
     double l;
     double r;
 }Area;
 Area area[1000];

 int cmp(const void *a,const void *b)
 {
     Area *A_a=(Area *)a;
     Area *A_b=(Area *)b;
     return (*A_a).l>(*A_b).l?1:-1;
 }

 int main()
 {
     int n,d,x,y,flag,case_num,num=1;
     while(cin>>n>>d&&!(n==0&&d==0))
     {
         flag=0;

         for(int i=0;i<n;i++)
         {
             cin>>x;
             cin>>y;
             if(y<=d&&flag==0)
             {
                 area[i].l=x-sqrt(double(d*d-y*y));
                 area[i].r=x+sqrt(double(d*d-y*y));
             }
             else flag=1;
         }

         if(flag==1) 
         {
             cout<<"Case "<<num<<": "<<-1<<endl;
             num++;
             continue;
         }
         qsort(area,n,sizeof area[0],cmp);

         double right=area[0].r;
         case_num=1;

         for(int i=1;i<n;i++)
         {
             if(area[i].l>right)
             {
                 case_num++;
                 right=area[i].r;
             }
             else if(area[i].r<right) right=area[i].r;
         }

         cout<<"Case "<<num<<": "<<case_num<<endl;
         num++;
     }
     return 0;
 }