Radar Installation（区间选点）

Radar Installation

Time Limit:1000MS     Memory Limit:10000KB     64bit IO Format:%lld & %llu

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

/*AC*/#include<vector>#include<cstdio>#include<algorithm>#include<cmath>using namespace std;struct data{    double l;    double r;};bool cmp(struct data a,struct data b){    if (a.r!=b.r)        return a.r<b.r;    return a.l>b.l;}int main(){    int n,r,count=1;    vector<struct data> v(1000);    while (scanf("%d%d",&n,&r))    {        if (n==0&&r==0)            break;        int i,x=0,y=0,flag=0;        for (i=0;i<n;i++)        {            scanf("%d%d",&x,&y);            if (y>r)                flag=-1;//这个条件千万不能漏，我第一次写就漏了这个条件            v[i].l=x-(double)sqrt(r*r-y*y);            v[i].r=x+(double)sqrt(r*r-y*y);        }        sort(v.begin(),v.begin()+n,cmp);        double t=v[0].r;        for (i=1;i<n&&flag!=-1;i++)        {            if (v[i].l>t)            {                t=v[i].r;                flag++;            }        }        if (flag!=-1)            flag+=1;        printf("Case %d: %d\n",count++,flag);    }    return 0;}