题解：Radar Installation（贪心）

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

在海岸线在放雷达，半径r，要求覆盖全部岛屿所用的最少雷达

以岛屿作圆，不以海岸线

贪心

#include
#include
#include
#include
#include
using namespace std;
struct node{
 double st,en;
 bool operator < (const node &x) const{
  return end) flag=1; //超出半径范围，不可能覆盖到 
   if(!flag)
  {
   double h=sqrt(d*d-y*y);
   a[i].st=x-h;a[i].en=x+h;//左，右端点 
  }
 }
 printf("Case %d: ",tip);
 if(flag) 
 {
  printf("-1\n");continue;
 }
   sort(a+1,a+n+1);//对右端点进行排序 
  int sum=1;
 double tmp=a[1].en;
  for(int i=1;i<=n;i++)
   if(a[i].st>tmp)
   {
    sum++;tmp=a[i].en; 
  }
 printf("%d\n",sum);
 }
 return 0;
}