POJ 1328 贪心

算法：

1.求出覆盖该岛的圆得区间, 將问题转换为求过出最少得点，保证每个区间至少有一个点。

2.按区间的左端排序

3.更新rad

#include <stdio.h>

#include <iostream>

#include <stdlib.h>

#include <string.h>

#include <algorithm>

#include <math.h>

using namespace std;



struct node

{

  double x, y; 

  bool operator < (const node& A) const

  {

     return x < A.x;     

  }

}px[10000], seg[10000];



double Circle1( double x, double y, double R)

{

       

   return x + sqrt( R * R - y * y );      

       

}





double Circle2( double x, double y, double R)

{

       

   return x - sqrt( R * R - y * y );      

       

}



int main( )

{

   int n, d, ans = 1;

   while( scanf("%d%d", &n, &d), n + d)

   {  

      int t = 0, num = 0;

      double ymax = 0;

      for( int i = 0; i < n; i++)

      {

        scanf("%lf%lf",&px[i].x, &px[i].y);

        if( px[i].y > ymax )

            ymax =  px[i].y;

      }

      if( ymax > d  )

      {

         printf("Case %d: %d\n", ans++, -1);

         continue;    

      }

      //求出圆心区间【X，Y】

      for( int i = 0; i < n; i++)

      {

         double x1 = px[i].x;

         double y1 = px[i].y;

         double Y = Circle1( x1, y1, d);

         double X = Circle2( x1, y1, d);

         seg[t].x = X;

         seg[t].y = Y;

         t++;

      }

      sort(seg, seg + t );
      double rad = seg[0].y;
      num = 1;

      for( int i = 1; i < t; i++)
      {
          if( rad - seg[i].x > 10e-7 ) 
          {
             if( seg[i].y - rad < 10e-7 )
         rad = seg[i].y; 
           
          }
          else if( seg[i].x - rad > 10e-7 )
          {
              rad = seg[i].y;
              num++;

          }
         

      }

      printf("Case %d: %d\n", ans++, num);  

   }

    

}