Description
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
大概题义就是要在x轴上放圆心，使得放最少的圆可以把所有点覆盖，输出圆的个数
#include <iostream>
#include <string.h>
#include <math.h>
#include <stdio.h>
#include <algorithm>
using namespace std;
#define Maxn 10010
struct Node{
double a,b;
}A[Maxn];
bool cmp(Node a,Node b){
if(a.a != b.a){
return a.a < b.a;
}else{
return a.b < b.b;
}
}
double wenbao(double R,double high){
return sqrt( pow(R,2.0) - pow(high,2.0) );
}
int main(){
int N;
double R;
double a,b;
int count = 1;
while(cin >> N >> R,N+R){
bool flag = true;
for(int i = 0; i < N; i++){
// cin >> a >> b;
scanf("%lf%lf",&a,&b);//用scanf,否则会超时
if(fabs(b) > R){
flag = false;
}else{
double temp = wenbao(R,b);
A[i].a = a - temp;
A[i].b = a + temp;
}
}
sort(A,A+N,cmp);
Node p = A[0];
int cnt = 1;
for(int i = 1; i < N; i++){
if(A[i].a > p.b){
cnt++;
p = A[i];
}else if(A[i].b < p.b){//请仔细思考
p = A[i];
}
}
if(flag){
printf("Case %d: %d\n",count++,cnt);
}else{
printf("Case %d: -1\n",count++);
}
}
}