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
想要按照最少的雷达站，先把x排序，如何判断下一个的最左边是不是在之前的雷达的最右边的右边，是就要重新按一个雷达站；
要注意，如果下一个的最右边在之前雷达的左边，说明安装的位置要在这里的最右边；
#include<iostream>
#include<stdio.h>
#include<stdlib.h>
#include<cmath>
#include<string.h>
#include<algorithm>
#define sf scanf
#define pf printf
#define cl clear()
#define pb push_back
#define mm(a,b) memset((a),(b),sizeof(a))
#include<vector>
const double pi=acos(-1.0);
typedef __int64 ll;
typedef long double ld;
const ll mod=1e9+7;
using namespace std;
int n,d,bits=1;
double wzl(int y)
{
double x=d*d-y*y;
return sqrt(x);
}
int a[1005],b[1005];
int main()
{
while(1)
{
mm(a,0);
mm(b,0);
sf("%d%d",&n,&d);
if(n==0)
return 0;
int temp=0;
for(int i=0;i<n;i++)
{
sf("%d%d",&a[i] ,&b[i] );
if(b[i]>d)
temp=1;
}
if(temp)
{
pf("Case %d: -1\n",bits++);
continue;
}
for(int i=0;i<n-1;i++)
for(int j=0;j<n-i-1;j++)
{
if(a[j]>a[j+1])
{
int w=a[j];
a[j]=a[j+1];
a[j+1]=w;
w=b[j];
b[j]=b[j+1];
b[j+1]=w;
}
}
int sum=1;
double last=a[0]+wzl(b[0]),left,right;
for(int i=1;i<n;i++)
{
left=a[i]-wzl(b[i]);
right=a[i]+wzl(b[i]);
if(left>last)
{
sum++;
last=right;
}else if(right<last)
{
last=right;
}
}
pf("Case %d: %d\n",bits++,sum);
}
}