Radar Installation---（贪心）

Radar Installation

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 115873		Accepted: 25574

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

分析：首先根据小岛的坐标计算出每座小岛对应海岸线上的范围。将每个小岛对应在海岸线上的范围进行排序，使得每个雷达范围的最小值进行递增。

对雷达范围进行贪心。。。

 #include<iostream>
 #include<cstdio>
 #include<cstring>
 #include<cmath>
 #include<algorithm>
 using namespace std;
 const int maxn=;
 #define INf 0x3f3f3f3f
 struct node{
     double l,r;
 }point[maxn];

 bool cmp(const node &a,const node &b){
     return a.l<b.l;
 }

 int main(){
     int n,d;
     int case1=;
     while(~scanf("%d%d",&n,&d)&&n){
         int flag=;
         for(int i=; i<n; i++ ){
             int x,y;
             cin>>x>>y;
             if(y>d){
                 flag=;
 //                break;
             }
             double p=sqrt((double)(d*d)-y*y);
             point[i].l=x-p;
             point[i].r=x+p;
         }
         printf("Case %d: ",++case1);
         if(flag){
             cout<<-<<endl;
             continue;
         }
         sort(point,point+n,cmp);
         int ans=;
         node tmp=point[];
         for( int i=; i<n; i++ ){
             if(tmp.r>=point[i].r) tmp=point[i];
             else if(tmp.r<point[i].l){
                 ans++;
                 tmp=point[i];
             }
         }
         cout<<ans<<endl;
     }
     return ;
 }