• ### 题目描述

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.

• ### 输入

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

• ### 输出

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.

• ### 样例输入

3 2
1 2
-3 1

2 1
1 2
0 2

0 0


Case 1: 2
Case 2: 1

• ### 思路

由于所有雷达全在x轴上，可把每个岛屿以d为圆心画圆，记录圆与x轴的两个交点存入pair中，把所有点对按右交点升序排列，判断后一个左交点是否小于前一个右交点即可。

• ### 代码

#include<iostream>
#include<map>
#include<set>
#include<queue>
#include<stack>
#include<vector>
#include<algorithm>
#include<string>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<cstdlib>
using namespace std;

vector<pair<double,double> > v;

int main()
{
int n,d,cnt=0;
while(cin>>n>>d)
{
cnt++;
if(n==0&&d==0)
return 0;
int maxx=-1;
for(int i=0;i<n;i++)
{
int x,y;
cin>>x>>y;
v.push_back(make_pair(x+sqrt(abs(1.0*d*d-y*y)),x-sqrt(abs(1.0*d*d-y*y))));
maxx=max(maxx,y);
}
if(maxx>d)
{
cout<<"Case "<<cnt<<": -1"<<endl;
continue;
}
sort(v.begin(),v.end());
int ans=1;
double tmp=v[0].first;
for(int i=1;i<n;i++)
{
if(v[i].second>tmp)
{
ans++;
tmp=v[i].first;
}
else if(v[i].first<tmp)
tmp=v[i].first;
}
cout<<"Case "<<cnt<<": "<<ans<<endl;
v.clear();
}
}