POJ 1328 Radar Installation 贪心

                                                      Radar Installation

 

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
代码 ：

#include<cstdio>
#include<iostream>
#include<cmath>
#include<algorithm>
#include<cstring>
#define M 1010
using namespace std ;
struct node { double  s , e ; } qe[M] ;
bool vi[M] ;
double ab( double  h , double y ){
   return sqrt( h * h - y * y ) ;
}
double ma( double a , double b ){
	if( a > b ) return a ;
	return b ;
}
int cmp( node a , node b ){
	return a.e < b.e ;
}
int main()
{
   int i , j  , n ,sum , k = 0 ;
   double u , v , m , max  ;
   while(scanf( "%d%lf" , &n , &m ) != EOF ){
	   if( n == 0 && m == 0.0 ) break ;
	   max = 0.0 ;k++ ;
	   memset( vi , 0 , sizeof(vi) ) ;
	   for( i = 1 ;i <= n ;i++){
		   scanf( "%lf%lf" , &u , &v ) ;
		   qe[i].s = u - ab( m , v ) ;
		   qe[i].e = u + ab( m , v ) ; // 把在能覆盖改点的左右端点 求出
		   max = ma( v , max ) ;
	   }
	    printf( "Case %d: " , k ) ;
	   if( max > m ) cout << "-1" << endl ; // 如果 最远的 不能覆盖 就无解
	   else {
           sort( qe + 1 , qe + n + 1 , cmp ) ; //  按右端点从小到大排序
		   for( i = 1 ;i <= n ;i++)
			   for( j = i + 1 ; j <= n ;j++){
				   if(qe[i].s >= qe[j].s )   // 如果被覆盖就不考虑
					   vi[j] = 1 ;
			   } 
			 //  for( i = 1 ;i <= n ;i++)
				  // printf( "%.2lf %.2lf\n" , qe[i].s , qe[i].e ) ;
			   sum = 0 ;
			   for( i = 1 ;i <= n ;i++)
				   if(vi[i] == 0 ){
					   max = qe[i].e ; 
					   sum++ ;
					   for( j = i + 1 ; j <= n ;j++){
						   if( qe[j].s <= max ) vi[j] = 1 ; // 覆盖
						   else break ;
					   }
				   }
				   cout << sum << endl ;
	   }
   }
}