Radar Installation

Time Limit: 1000ms, Special Time Limit:2500ms, Memory Limit:32768KB

Total submit users: 562, Accepted users: 500

Problem 10023 : No special judgement

Problem 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. 

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 21 2-3 12 11 20 20 0

Sample Output

Case 1: 2Case 2: 1

Problem Source

Bei jing 2002

解题思路：

radar installation

在海里有很多小岛，每个小岛位置由x,y坐标确定。而我们只能在海岸线上建雷达站，同时给定雷达站的最大覆盖距离d（以雷达站为圆心，d为半径），要求所有小岛都要被雷达站覆盖，求最少的雷达站数目，如果没有解决方案，输出-1。

 

贪心策略1：让某个雷达站覆盖尽量多的小岛？--》不正确

贪心策略2：从左往右建雷达站：最左边的雷达站需要覆盖最左边的小岛，同时位置尽量右靠。依次重复。

1. 如何确定雷达可能位置：以小岛为圆心，画圆，与海岸线的右交点为建雷达站点。

2. 对于所有小岛，求出左右交点，从左至右，以第一个右交点为基准，

如果下一海岛的左交点小于或等于该右交点，则此海岛能够被该雷达站覆盖，

不过这里要注意，如果出现该海岛的右交点小于或该右交点的情况，基准应该换成该海岛的右交点；

如果下一海岛的左交点大于该右交点，则需要新建雷达站，雷达站数++，同时以这一海岛的右交点为基准，重复上步骤；

 

代码（C++）:

#include
     
      #include 
      
       #include
       
        using namespace std;struct node{    double left,right;//以每个岛为圆心，与x轴的左右交点 }island[1001];//1001个实例 //sort的比较函数 bool cmp(node a,node b){    return a.left
        
         >n>>d && n!=0){        CaseNum++;         flag=0;                 //输入，并记录island.left 和island.right         for(int i=0;i
         
          >x>>y;            //如果小岛到x轴的距离y大于d，则没有解决方案            if(y>d) {                flag=1;//没有解决方案             }            //记录以每个小岛为圆心，d为半径的圆与x轴的左右交点             island[i].left=x-sqrt(d*d-y*y);            island[i].right=x+sqrt(d*d-y*y);        }                //没有解决方案         if(flag==1) {            cout<<"Case "<
          
           <<": -1"<
           
            tmp){ count++; tmp=island[i].right; } else if(island[i].right