Radar Installation

题目：

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

 方法解析：

大致可以分为两种情况：

1.当所给点的y轴坐标有y>d（雷达影响半径）的情况时，则是无解的情况。

2.当所有点的y轴坐标都不大于d时：

    以每一个点为圆心，d为半径画圆。每一个圆都会与x有两个交点，即左点（a表示）和右点（b 表示）（只有一个的也看成有值相等的两个点）。按右点的x轴坐标（即bx）把所有b点所对应的岛屿（输入点）从小到大排序。从左至右，以排好的b点为雷达圆心。每画一个雷达圆，标记ax<=bx的a点所对应的岛屿点，那么下一个圆的圆心坐标为第一个没有标记的岛屿所对应的b点。当有新点（未标记的点）被标记时，雷达个数加一（是处理同一个雷达圆时所遇到的新点，不重复计数）。

    ............自己都有点绕晕了......= =

代码如下：

#include<iostream>
#include<cstring>
#include<cmath>
#include<cstdio>
using namespace std;

class Num
{
public:
    double x,y,a,b,p;
};

#define M 1000

double ffind(double y,double r)
{
    return sqrt(r*r-y*y);
}

int main()
{
    Num id[M];
    Num t;
    double m,r;
    int step=0;
    while(scanf("%lf%lf",&m,&r)!=EOF)
    {
        if(m==0 && r==0) break;
        step++;
        int i,j,number=0,k=0;
        for(i=0;i<m;i++)
        {
            id[i].x=0;
            id[i].y=0;
            id[i].a=0;
            id[i].b=0;
            id[i].p=0;
        }
        for(i=0;i<m;i++)
        {
            scanf("%lf%lf",&id[i].x,&id[i].y);
            if(id[i].y>r) number=-1;
            id[i].a=id[i].x-ffind(id[i].y,r);
            id[i].b=id[i].x+ffind(id[i].y,r);
        }
        for(i=0;i<m;i++)
            for(j=i;j<m;j++)
        {
            if(id[j].b<id[i].b)
            {
                t=id[j];
                id[j]=id[i];
                id[i]=t;
            }
        }
        for(i=0;i<m && number>=0;i++)
        {
            if(id[i].p==0)
            {
            for(j=0;j<m;j++)
            {
                if(id[j].p==0 && id[j].a<=id[i].b)
                {
                    id[j].p=1;
                    k++;
                }
            }
            }
            if(k>0){ number++;}
            k=0;
        }
       printf("Case %d: %d\n",step,number);
    }
    return 0;
}