Radar Installation

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<tex2html_verbatim_mark> distance, so an island in the sea can be covered by a radius installation, if the distance between them is at most d<tex2html_verbatim_mark> .

We use Cartesian coordinate system, defining the coasting is the x<tex2html_verbatim_mark> -axis. The sea side is above x<tex2html_verbatim_mark> -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<tex2html_verbatim_mark> - y<tex2html_verbatim_mark>coordinates.

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The input consists of several test cases. The first line of each case contains two integers n<tex2html_verbatim_mark>(1n1000)<tex2html_verbatim_mark> and d<tex2html_verbatim_mark> , where n<tex2html_verbatim_mark> is the number of islands in the sea and d<tex2html_verbatim_mark> is the distance of coverage of the radar installation. This is followed by n<tex2html_verbatim_mark> 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.

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.

3 2
1 2
-3 1
2 1

1 2
0 2

0 0

Case 1: 2
Case 2: 1

思路：该题题意是为了求出能够覆盖所有岛屿的最小雷达数目，每个小岛对应x轴上的一个区间，在这个区间内的任何一个点放置雷达，则可以覆盖该小岛，区间范围的计算用[x-sqrt(d*d-y*y),x+sqrt(d*d-y*y)];这样，问题即转化为已知一定数量的区间，求最小数量的点

#include<iostream>
#include<cmath>
#include<algorithm>
using namespace std;
int n,d;

class M{
public :
    double x,y;
    bool operator<(M c)const{
        return y<c.y;
    }
    void fun(){
        double t=d*d-y*y;
        y=sqrt(t)+x;
        x=x-sqrt(t);
    }
};
M m[1005];
bool judge(double x,M a){
    return x>=a.x&&x<=a.y;
}
int main(){
        int k=0;
    while(cin>>n>>d&&n&&d){
            int x,y;int ok=1;
            int n1=n;
        for(int i=0;i<n;i++){
            cin>>x>>y;
            if(y>d){ok=0;n1--;continue;}
            else {
                m[i].x=x;
                m[i].y=y;
                m[i].fun();
            }
        }
        n=n1;
        if(ok==0){cout<<"Case "<<++k<<": -1"<<endl;;continue;}
        sort(m,m+n);
        int sum=0;
        for(int i=0;i<n;i++){
            double t=m[i].y;
            while(i+1<n&&judge(t,m[i+1])){
                i++;
            }
            sum++;
        }
        cout<<"Case "<<++k<<": "<<sum<<endl;
    }
return 0;
}

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