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
AC代码：
//贪心，主要弄清排序原则，，遍历的过程
//用代码实现自己的想法才行。。
#include<iostream>
#include<algorithm>
#include<cstdio>
#include<math.h>
using namespace std;
struct stu
{
double l,r;//能被覆盖的所有圆中，左端点和右端点
}I[1005];
bool cmp(stu a,stu b)
{
//右端点从小到大排序，右端点相同时，按照左端点从大到小排序
if(a.r==b.r)
return a.l>b.l;
return a.r<b.r;
}
int main()
{
int k=1;
int n,d;
double x,y;
while(cin>>n>>d && n)
{
int flag=0;
for(int i=0;i<n;i++)
{
cin>>x>>y;
I[i].l=x-sqrt(d*d-y*y);
I[i].r=x+sqrt(d*d-y*y);
if(d<y || d<0)//不够，或者半径<0
flag=1;
}
if(flag)
printf("Case %d: -1\n",k++);
else
{
sort(I,I+n,cmp);
int ans=1;
double post=I[0].r;//初始，右端点
for(int i=1;i<n;i++)
{
if(I[i].l>post)//超过上一个范围的最大值了
{
ans++;
post=I[i].r;//不断更新右端点
}
}
printf("Case %d: %d\n",k++,ans);
}
}
return 0;
}
具体思路见https://blog.csdn.net/qq_42891420/article/details/89086025