poj 1328 Radar Installation

贪心：

题意：给定小岛的位置，问最小安装雷达数，雷达有一个半径，覆盖小岛。我竟然看错了题意，雷达都安装在x轴上，我竟然不知道。

分析：以小岛做一个圆，圆与x轴有2个交点a,b在区间[a,b]选点的问题，就转换区间选点问题，在小白书1的153页，先把区间b从小到大排序，取最后一个点。

Radar Installation

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 55095		Accepted: 12415

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

Sample Output

Case 1: 2
Case 2: 1

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<cstring>
 4 #include<algorithm>
 5 #include<cmath>
 6 using namespace std;
 7 struct node
 8 {
 9     double x,y;
10 }a[1005];
11 bool cmp(node d,node c)
12 {
13     return d.y<c.y;
14 }
15 int main()
16 {
17     int n,i,j,t=1,flag,ans;
18     double d,r,w;
19     while(~scanf("%d%lf",&n,&r))
20     {
21             ans=1;
22            flag=0;
23         if(n==0&&r==0)
24              break;
25         for(i=0;i<n;i++)
26         {
27             scanf("%lf%lf",&w,&d);
28             if((r*r-d*d)<0)
29                 flag=1;
30              a[i].x=w-sqrt(r*r-d*d);
31               a[i].y=w+sqrt(r*r-d*d);
32         }
33           if(flag||r<=0)
34            {
35                 printf("Case %d: -1
",t++);
36                 continue;
37            }
38         sort(a,a+n,cmp);
39         double temp=a[0].y;
40         for(i=1;i<n;i++)
41         {
42             if(a[i].x>temp)
43               {
44                   ans++;
45                   temp=a[i].y;
46               }
47         }
48         printf("Case %d: %d
",t++,ans);
49     }
50     return n;
51 }