1.Link：

http://poj.org/problem?id=1328

http://bailian.openjudge.cn/practice/1328/

2.Content：

Radar Installation

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 52833		Accepted: 11891

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

Beijing 2002

3.Method：

（1）求出每个岛能够安装灯塔的区域，用结构体表示Lines

（2）根据最右快排，这里要尤其注意double类型的compare怎么写

（3）贪心算法求最小灯塔数量

4.Code：

#include <iostream>
#include <cstdio>
#include <cmath>
#include <cstdlib>

using namespace std;

struct LINE
{
    double l;
    double r;
};

int cmp(const void *a,const void *b)
{
    LINE line1 = *((LINE *)a);
    LINE line2 = *((LINE *)b);
    /*if(line1.r == line2.r)
    {
        if(line1.l == line2.l) return 0;
        return line1.l > line2.l ? 1 : -1;    
    }    
    else return line1.r > line2.r ? 1 : -1;*/
    if(line1.r == line2.r) return 0;
    else return line1.r > line2.r ? 1 : -1;
}

int main()
{
    //freopen("D://input.txt", "r", stdin);

int i;

int n,d;
    int count = 1;
    int x,y;
    int flag;
    double ins;
    
    cin >> n >> d;
    while(n!= 0 || d != 0)
    {

LINE *lines = new LINE[n];

flag = 1;
        for(i = 0; i < n; ++i)
        {
            cin>>x>>y;
            if(d < y) flag = 0;
            else
            {
                ins = sqrt(d * d - y * y);
                lines[i].l = x - ins;
                lines[i].r = x + ins;
            }
        }

if(flag == 0) cout<<"Case "<< (count++) <<": -1"<<endl;
        else
        {
            qsort(lines,n,sizeof(LINE),cmp);
            
            /*for(i = 0; i < n; ++i)
            {
                cout << lines[i].l << " " << lines[i].r << endl;
            }*/
            
            int num = 1;
            double max_r = lines[0].r;
            for(i = 1; i < n; ++i)
            {
                if(max_r < lines[i].l)
                {
                    num++;
                    max_r = lines[i].r;
                }
            }
            cout << "Case " << (count++) << ": " << num <<endl;
        }
                
        delete [] lines;
        
        cin >> n >> d;
    }

//fclose(stdin);
    return 0;
}

5：Reference：

http://blog.sina.com.cn/s/blog_48f85e1d0100nslz.html