Radar Installation POJ - 1328（贪心）

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

题意：N座岛屿，坐标用xi，yi表示，岛屿都位于x轴上。在x轴上建造雷达，每个雷达的的探测范围是以【x，0】为圆心，半径为R的圆（下半圆一看就没用），问最少几个雷达可以探测全部岛屿？

思路：我们把每个岛屿为圆心，做出x轴上雷达的可行建造区域，那么问题就变成用最小的雷达覆盖最多的区域（每个区域一个雷达）。
我们先将区域按左边界递增排序，每取到一个区间，就记录在这个区间右边界pos，如果下个区间的左边界>pos，说明这个雷达无法探测，需要新建雷达，
否则就pos = min（pos，r），就是让原有的雷达去探测它，但是雷达需要移至两个区间的最小右边界处（交集嘛）。

①因为要用比较左边界和上一次区间的右边界，所以一定是按照左边界排序的。（这样处理后面的区域和之前区域没啥关系，只和pos有关）
置于为什么pos取右边界，因为右边界代表了雷达最远可放置位置（极值），其包含了区间内可行取值了。（决策包容性？）

②当然如果你反过来，右边界排序，你就需要比较右边界和上一次的左边界也行。

 #include<cstdio>
 #include<iostream>
 #include<algorithm>
 #include<math.h>
 using namespace std;

 int n,r;
 struct Node
 {
     double l,r;
 } L[];

 double calc(int y,int r)
 {
     return sqrt(r*r-y*y);
 }

 int dcmp(double a,double b)
 {
     if(fabs(a-b) < 1e-)
         return ;
     else
         return a-b > ?:-;
 }
 bool cmp(Node a,Node b)
 {
     return dcmp(a.l,b.l)<;
 }
 int main()
 {
     int cas = ;
     while(~scanf("%d%d",&n,&r) && (n || r))
     {
         int ans=;
         for(int i=; i<=n; i++)
         {
             double x,y;
             scanf("%lf%lf",&x,&y);
             if(dcmp(y,r) > )
             {
                 ans = -;
                 continue;
             }
             double add = calc(y,r);
             L[i].l = x-add;
             L[i].r = x+add;
         }
         if(ans == -){printf("Case %d: -1\n",++cas);continue;}
         sort(L+,L++n,cmp);
         double pos = -0x3f3f3f3f3f3f3f;
         for(int i=; i<=n; i++)
         {
             if(dcmp(L[i].l,pos)>)
             {
                 ans++;
                 pos = L[i].r;
             }
             else
             {
                 pos = min(L[i].r,pos);
             }
         }
         printf("Case %d: %d\n",++cas,ans);
     }
 }