Radar Installation
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.
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
题目大意：x轴上面有N个P点，给你一个d表示半径，问在x轴上需要几个半径为d的点才能将所有P点覆盖。
题解：贪心算法。算出p点到x轴上距离为d的区间【x1,x2】然后按照x2排序，查区间重叠的部分。
具体代码如下：
#include<iostream>
#include<algorithm>
#include<cmath>
using namespace std;
typedef struct{
int X,Y;
double x1,x2;
}point;
bool cmp(point a,point b)
{
return a.x2<b.x2;
}
int main()
{
int n,num=0;
double d;
while(cin>>n,cin>>d,n&&d)
{
num++;
int i;
point *a;
a=new point[n+1];
int flag=0;
for(i=0;i<n;i++)
{
cin>>a[i].X>>a[i].Y;
if(d*d-a[i].Y*a[i].Y*1.0>=0.0)
{
a[i].x1=a[i].X*1.0-sqrt(d*d-a[i].Y*a[i].Y*1.0);
a[i].x2=a[i].X*1.0+sqrt(d*d-a[i].Y*a[i].Y*1.0);
}else{
flag=1;
break;
}
}
if(flag==1)
cout<<"Case "<<num<<": -1"<<endl;
else
{
sort(a,a+n,cmp);
double p=a[0].x2;
int final=1;
for(i=0;i<n;i++)
{
if(a[i].x1<=p)
{
continue;
}else{
final++;
p=a[i].x2;
}
}
cout<<"Case "<<num<<": "<<final<<endl;
}
}
return 0;
}