区间重叠问题 （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

大意： 海岸线的上方是islands，海岸线上有雷达，给出islands坐标，能够把所有岛检测到而且需要最少数量的雷达，求这个最小数量。
解题思路：①给的points（islands）是确定的，而雷达是不确定的，我们可以以每一个islands为圆心以输入r为半径做圆，因为雷达在海岸线上（x轴），如果这个圆与x轴相交，设交点为x1，x2（由圆的公式或者勾股定理加已知条件，可以求出两点坐标），那么在这一段x1-x2中，必然存在一个雷达可以检测到这个岛（距离小于半径），那么，我们此时做第二个点的圆（假设我们的点已经排好序，按x的大小，如果x相等，把y小的放在前面，也可以不考虑y，自己分析），假设也会与x轴相交（如果不相交，肯定输出-1，因为没有雷达会检测到，不存在这样的雷达），设交点为x3，x4，如果区间x1—x2与区间x3—x4相交，那么存在一个雷达会同时检测到这两个点，如果不相交，那么就会需要两个雷达，依次类推，但需要注意一点，有可能第二个点的x4<第一个点的x2，这样第二个区间被第一个区间完全包含，这样我们在判断时会产生一个区间范围缩小的问题（具体看代码实现）。
②以雷达为考虑中心，把小岛考虑在某个雷达监测的最边上，这样岛左边与右边各有一个极限雷达，两个雷达之间必然存在一个雷达，包括两个极限雷达可以检测到这个岛的存在，也是区间问题，作为一种参考思路，不做详述。

一下是第一种方法的AC代码，仅供参考，新手，难免代码有缀（一开始自己有思路，但是没考虑区间包含，参考了其他博主代码，做了代码优化）。
#include
<iostream>
#include<math.h>
#include
<algorithm>
using namespace
std;
typedef struct points{
double x;
double y;}type;bool
cmp(type a,type b)
{

   if (a.x==b.x)

   return a.y<b.y;

   return a.x<b.x;

}
int
main(){

type *pt,*xd;
double x1,x2,border;
int i,j,k,n,count=1,r,flag=1,flag1=0;
while (scanf("%d%d",&n,&r))

{
if (n==0&&r==0)
break;

pt=new type
;

xd=new type
;
for (i=0;i<n;i++)

{
cin>>pt[i].x>>pt[i].y;

  if (pt[i].y>r||r<0||pt[i].y<0)

flag1=1;

    }

    if (flag1)

{

     cout<<"Case"<<" "<<flag<<":"<<" "<<"-1"<<endl;

flag1=0;

flag++;

}

    else

    {

sort (pt,pt+n,cmp);
for (i=0;i<n;i++)

{

xd[i].x=pt[i].x-sqrt(r*r-pt[i].y*pt[i].y);

xd[i].y=pt[i].x+sqrt(r*r-pt[i].y*pt[i].y); }border=xd[0].y;

    for (i=1;i<n;i++)

    {

      if (border>xd[i].y)

       border=xd[i].y;

       else

       {

       if (xd[i].x>border)

       {

border=xd[i].y;

       count++;

         }

   }

}
cout<<"Case"<<" "<<flag<<":"<<" "<<count<<endl;

count=1;

flag++;

    }

}
return 0;

}