NYOJ-287 Radar

Radar

时间限制：1000 ms  |  内存限制：65535 KB

难度：3

描述

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.

 

输入

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

输出

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.

样例输入

3 21 2-3 12 11 20 20 0

样例输出

Case 1: 2Case 2: 1

算出每个岛到坐标左右最远的距离，转化为区间选点

不知道为啥，加了无满足条件输出-1不给过……      

01.#include<iostream>

02.#include<algorithm>

03.#include<cmath>

04.using namespace std;

05.struct node

06.{

07.double x,y;

08.}no[1010];

09.bool cmp(node p,node q)

10.{

11.return p.y<q.y;

12.}

13.int main()

14.{

15.int n,sum=1;

16.double b,l,r;//max=-10000.0;

17.while(cin>>n>>b&&n&&b)

18.{

19.for(int i=0;i<n;i++)

20.{  

21.cin>>l>>r;

22.no[i].x=l-sqrt(b*b-r*r);

23.no[i].y=l+sqrt(b*b-r*r);

24.//max=max>r?max:r;          

25.}  

26.sort(no,no+n,cmp); 

27.int count=1;

28.double v=no[0].y;

29.for(int i=1;i<n;i++)

30.{

31.if(no[i].x>v)

32.{

33.v=no[i].y;

34.count++;

35.}              

36.}

37.//if(max>b)

38.//  cout<<"Case "<<sum++<<": "<<-1<<endl;

39.//else

40.cout<<"Case "<<sum++<<": "<<count<<endl;

41.}

42.return 0;

43.}