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
解题思路：该题题意是为了求出能够覆盖所有岛屿的最小雷达数目，每个小岛对应x轴上的一个区间，在这个区间内的任何一个点放置雷达，则可以覆盖该小岛，区间范围的计 算用[x-sqrt(d*d-y*y),x+sqrt(d*d-y*y)];这样，问题即转化为已知一定数量的区间，求最小数量的点，使得每个区间内斗至少存在一个点。每次迭代对于第一个区间，选 择最右边一个点， 因为它可以让较多区间得到满足， 如果不选择第一个区间最右一个点（选择前面的点）， 那么把它换成最右的点之后，以前得到满足的区间， 现在仍然 得到满足， 所以第一个区间的最右一个点为贪婪选择， 选择该点之后， 将得到满足的区间删掉，进行下一步迭代， 直到结束。
1 #include <iostream>
2 #include <cstring>
3 #include <cstdio>
4 #include <algorithm>
5 #include <math.h>
6 using namespace std;
7 struct node
8 {
9 double l,r;
10 };
11 node a[1005];
12 bool cmp(node a,node b)
13 {
14 return a.r<b.r;
15 }
16 int main()
17 {
18 int n,k=1;
19 int flag;
20 double d;
21 while(cin>>n>>d)
22 {
23 flag=1;
24 if(n==0&&d==0)
25 {
26 break;
27 }
28 double x,y;
29 for(int i=0;i<n;i++)
30 {
31 cin>>x>>y;
32 if(!flag)
33 {
34 continue;
35 }
36 if(y>d)
37 {
38 flag=0;
39 continue;
40 }
41 a[i].l=x-sqrt(d*d-y*y);
42 a[i].r=x+sqrt(d*d-y*y);
43 }
44 printf("Case %d: ", k++);
45 if(!flag)
46 {
47 printf("-1\n");
48 continue;
49 }
50 sort(a,a+n,cmp);
51 double temp=-10000;
52 int ans=0;
53 for(int i=0;i<n;i++)
54 {
55 if(temp<a[i].l)
56 {
57 ans++;
58 temp=a[i].r;
59 }
60 }
61 cout << ans << endl;
62 }
63
64 return 0;
65 }
原文地址：https://www.cnblogs.com/SoulSecret/p/8425976.html