Radar Installation

Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 20000/10000K (Java/Other)

Total Submission(s) : 54   Accepted Submission(s) : 28

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

题解：先转化为区间点，再排序，区间找点；

代码：

 1 #include<stdio.h>
 2 #include<math.h>
 3 #include<algorithm>
 4 using namespace std;
 5 struct Node{
 6     double s,e;
 7 };
 8 int n,d,k;
 9 Node area[1010];
10 int cmp(Node a,Node b){
11     return a.e<b.e;
12 }
13 int change(int x,int y){
14     if(y>d)return 0;
15     double a,b,m=sqrt(d*d-y*y);
16     a=x-m;b=x+m;
17     area[k].s=a;area[k].e=b;k++;
18     return 1;
19 }
20 int main(){int t,x,y,flot,temp,num,l=0;
21     while(scanf("%d%d",&n,&d),n||d){k=0;flot=1;temp=0;num=1;l++;
22             for(int i=0;i<n;i++){
23                 scanf("%d%d",&x,&y);
24                 t=change(x,y);
25                 if(!t)flot=0;
26             }
27             sort(area,area+k,cmp);
28             for(int i=0;i<k;i++){
29                 if(area[i].s>area[temp].e)temp=i,num++;
30             }
31             printf("Case %d: %d\n",l,num);
32     }
33     return 0;
34 }