Radar Installation

1328 poj

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 58386		Accepted: 13155

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

 

先将之转化为各区间 接下来就是区间选点问题了 排序后选最后一个点就OK；

 1 #include<stdio.h>

 2 #include<string.h>

 3 #include<algorithm>

 4 #include<math.h>

 5 using namespace std;

 6 struct node{

 7     double x;

 8     double y;

 9     double l;

10     double r;

11 }island[1005];

12 

13 bool cmp(node a,node b)

14 {

15     if(a.r!=b.r)

16         return a.r<b.r;

17     else

18         return a.l>b.l;

19 }

20 

21 int main()

22 {

23     int n,i,j,k,t=0,flg,num;

24     double d,local;

25     while(scanf("%d %lf",&n,&d)!=EOF)

26     {

27         flg=1;t++;num=1;

28         if(n==0 && d==0)

29         {

30             break;

31         }

32         for(i=1;i<=n;i++)

33         {

34             scanf("%lf %lf",&island[i].x,&island[i].y);

35             if(island[i].y>d)

36                 flg=0;

37         }

38         if(flg==0)

39         {

40             printf("Case %d: -1\n",t);

41             continue;

42         }

43         for(i=1;i<=n;i++)

44         {

45             island[i].l=island[i].x-sqrt(d*d-island[i].y*island[i].y);

46             island[i].r=island[i].x+sqrt(d*d-island[i].y*island[i].y);

47         }

48         sort(island+1,island+n+1,cmp);

49         local=island[1].r;

50         for(i=2;i<=n;i++)

51         {

52             if(island[i].l<=local)

53                 continue;

54             else

55             {

56                 num++;

57                 local=island[i].r;

58             }

59         }

60         printf("Case %d: %d\n",t,num);

61     }

62     return 0;

63 }

View Code