poj 1328 Radar Installation（贪心）

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 (<=n<=) 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 :
Case : 

Source

Beijing 2002

贪心。

一开始的思路：

图中ABCD为海岛的位置。假设本题半径为2（符合坐标系），那么A点坐标为（1，1）以此类推。

在题中，记录每个点的坐标，并把一个点新加一个标记变量以标记是否被访问过。

首先以A为圆心，r为半径做圆，交x轴与E1（右）点，做出如图中绿色虚线圆。然后以E1为圆心，半径为r做圆。看此时下一点（B）是否在该圆中。如果在，那么将该点标记变量变为true，再判下一点（C），如果不在那么就新增一个雷达。

后来，换了思路，存储图中红色圆与X轴的交点。

还是原来的贪心思路，仍然先排序，但排序的准则是按红色圆与x轴左交点的先后顺序。

如图，如果B点圆（且如此称呼）的左交点（J1点）在A点圆右交点（E1）左侧，那么B点一定涵盖在A点圆内部。再如图，如果D点圆的左交点（图中未标示）在A点圆的右交点（未标示）右，则D点不在A点圆中。

故，有如下代码：

 #include<iostream>
 #include<cstdio>
 #include<cstring>
 #include<algorithm>
 #include<stdlib.h>
 #include<cmath>
 using namespace std;
 struct Node
 {
     double left,right;
 }p[];
 bool cmp(Node a,Node b)
 {
     return a.left<b.left;
 }
 int main()
 {
     int n;
     double r;
     int ac=;
     while(scanf("%d%lf",&n,&r)== && (n || r))
     {
         int f=;
         for(int i=;i<n;i++)
         {
             double a,b;
             scanf("%lf%lf",&a,&b);
             if(b>r)
             {
                 f=;
             }
             else
             {
                 p[i].left=a-sqrt(r*r-b*b);
                 p[i].right=a+sqrt(r*r-b*b);
             }
         }
         printf("Case %d: ",++ac);
         if(f==)
         {
             printf("-1\n");
             continue;
         }
         sort(p,p+n,cmp);
         int ans=;
         Node tmp=p[];
         for(int i=;i<n;i++)
         {
             if(p[i].left>tmp.right)
             {
                 ans++;
                 tmp=p[i];
             }
             else if(p[i].right<tmp.right)
             {
                 tmp=p[i];
             }
         }
         printf("%d\n",ans);

     }
     return ;
 }