Radar Installation

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.   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

 

Source

PKU

解题思想：输入各个点；计算出每一个点的圆心在x轴上的范围，然后以圆心范围的前头为标准进行升序排序，让后利用贪心算法进行统计所需雷达的个数；

解题代码：

#include<iostream>
#include<cmath>
#include<cstdio>
#include <algorithm>
using namespace std;
const int Max(1005);
struct data
{
    double x1;
    double x2;
}point[Max];      //定义圆心范围的结构体,并且同时定义一个圆心范围的结构体数组；
int cmp(data a, data b)  //定义sort（快排函数）的比较函数；
{
    return (a.x1-b.x1) < 10e-7;
}

int main()
{
    int step=1;
    while(1)
    {
        int flag=0;         //标志位，表示是否有一个点无论如何都不能被检测到；
        int n,d;
        cin>>n>>d;
        if(n==0&&d==0)
            break;
        int a,b;
        for(int i=0;i<n;i++)  // 注意这边i不能从1开始，因为后面用到了sort排序
        {                    //输入各个点，并计算圆心范围；
            cin>>a>>b;
            if(!flag&&(b<= d))
            {
                double temp = sqrt(double(d*d-b*b));  //由点处理出圆心的范围
                point[i].x1 = a-temp;
                point[i].x2 = a+temp;
            }
            else
                flag=1;

        }
        if(flag==1)//存在一个点不可能被检测到；
        {
            cout<<"Case "<<step<<": "<<"-1"<<endl;
            step++;
        }
        else
        {
            int sum=1;
            sort(point, point+n, cmp);
            double temp=point[0].x2;
            for(int i=1;i<n;i++)
            {
                if(point[i].x1-temp>10e-7)
                {
                    sum++;
                    temp=point[i].x2;
                }
                else
                {
                    if(point[i].x2-temp< 10e-7)  //当走到的点的右范围比原来的右范围小，即temp相应的改
                    temp = point[i].x2;
                }
            }
            cout<<"Case "<<step<<": "<<sum<<endl;
            step++;
        }
    }
    return 0;

}