Radar Installation

Problem 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

PKU

解题思想：输入各个点；计算出每一个点的圆心在x轴上的范围，然后以圆心范围的前头为标准进行升序排序，让后利用贪心算法进行统计所需雷达的个数；

解题代码：

 1 #include<iostream>
 2 #include<cmath>
 3 #include<cstdio>
 4 #include <algorithm>
 5 using namespace std;
 6 const int Max(1005);
 7 struct data
 8 {
 9     double x1;
10     double x2;
11 }point[Max];      //定义圆心范围的结构体,并且同时定义一个圆心范围的结构体数组；
12 int cmp(data a, data b)  //定义sort（快排函数）的比较函数；
13 {
14     return (a.x1-b.x1) < 10e-7;
15 }
16 
17 int main()
18 {
19     int step=1;
20     while(1)
21     {
22         int flag=0;         //标志位，表示是否有一个点无论如何都不能被检测到；
23         int n,d;
24         cin>>n>>d;
25         if(n==0&&d==0)
26             break;
27         int a,b;
28         for(int i=0;i<n;i++)  // 注意这边i不能从1开始，因为后面用到了sort排序
29         {                    //输入各个点，并计算圆心范围；
30             cin>>a>>b;
31             if(!flag&&(b<= d))
32             {
33                 double temp = sqrt(double(d*d-b*b));  //由点处理出圆心的范围
34                 point[i].x1 = a-temp;
35                 point[i].x2 = a+temp;
36             }
37             else
38                 flag=1;
39 
40         }
41         if(flag==1)//存在一个点不可能被检测到；
42         {
43             cout<<"Case "<<step<<": "<<"-1"<<endl;
44             step++;
45         }
46         else
47         {
48             int sum=1;
49             sort(point, point+n, cmp);
50             double temp=point[0].x2;
51             for(int i=1;i<n;i++)
52             {
53                 if(point[i].x1-temp>10e-7)
54                 {
55                     sum++;
56                     temp=point[i].x2;
57                 }
58                 else
59                 {
60                     if(point[i].x2-temp< 10e-7)  //当走到的点的右范围比原来的右范围小，即temp相应的改
61                     temp = point[i].x2;
62                 }
63             }
64             cout<<"Case "<<step<<": "<<sum<<endl;
65             step++;
66         }
67     }
68     return 0;
69 
70 }

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