poj1328——Radar Installation（贪心）-白红宇

poj1328——Radar Installation（贪心）

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发布时间：2019-05-10

本文共 2108 字，大约阅读时间需要 7 分钟。

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

#include 
   
    #include 
    
     #include 
     
      #include 
      
       #include 
       
        //#include 
        
         #include 
         
          #include 
          
           #include 
           
            #include 
            #include 
             
              #define MAXN 1010#define mod 2012#define INF 0x3f3f3f3fusing namespace std;struct Node{ double left,right;};Node p[MAXN];bool cmp(Node a,Node b){ return a.left
              
               >n>>d&&(n||d)) { bool flag=false; for(int i=0; i
               
                >x>>y; p[i].left=x-sqrt(d*d-y*y); p[i].right=x+sqrt(d*d-y*y); if(fabs(y)>d) flag=true; } cout << "Case " << cnt++ << ": "; if(flag) cout<<"-1"<
                
                 tmp.right) { ans++; tmp=p[i]; } else if(p[i].right

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