POJ 1328 Radar Installation（贪心）

Radar Installation

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 89909 Accepted: 20195

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 21 2-3 12 11 20 20 0

Sample Output

Case 1: 2Case 2: 1

Source

Beijing 2002

思路：无解的情况显然很容易判定，只要有一个岛屿到海岸线的距离y>d，那么就无解。对每个岛屿，考虑能监控该岛屿的位置，显然是以该岛屿为圆心，d为半径的圆，与海岸线的两个交点的线段（可能退化）。把所有线段找出来，并按右端排序，若右端相同，左端大的排在右边。贪心策略：从未选择过的线段中选择一个右端点在最左边的线段，在该线段的右端点安装雷达。对进行过贪心选择的区间[a,b]标记覆盖标志，同时对其他的未访问过的区间且左端点坐标小于b的区间也标志覆盖标志，说明在b处安装雷达也可以监控到此岛屿。

代码：

bool cmp(Point a,Point b){         if(a.y!=b.y)                   return a.y<b.y;         else                   return a.x<b.x;}void init(){         for(int i=0;i<n;i++)         {                  cin>>point[i].x>>point[i].y;                  if(d>=point.y)                  {                            point[i].x=point[i].x-sqrt(double(d)*d-point[i].y*point[i].y);                            point[i].x=point[i].x+sqrt(double(d)*d-point[i].y*point[i].y);                  }                  else                  {                            printf("-1\n");                            return ;                  }          }}void solve(){          sort(point,point+n,cmp);          memset(visit,0,sizeof(visit));          int ans=0;          for(int  i=0;i<n;i++)          {                   if(!visit[i])                   {                            visit[i]=1;                            ans++;                            for(int j=i+1;j<n;j++)                                       if(!visit[i]&&point[j].x<=point[i].y)//未访问且左端点小于当前右端点                                                 visit[j]=1;                     }           }}