On an 1 × 1 grid,partition the length and width into 50 parts,forming 2500 unit grids with the same size.Now,given n circles with equal radius R on the grid.Find the total number of unit grids which are not covered by any of a circle.
　　A unit grid is covered by a circle,means that the distance between the center of the unit grid and the center of circle is less or equal R.
							

　　There will be multiple test cases.Each line of input contains aInteger n (1≤n≤100) and a decimal R (0.01≤R≤0.1).Follows n lines,each line describe the center of circle i,include two decimal xi, yi (0≤xi, yi≤1).
							

　　For each test case output a line,output the number of the points which are not covered.
							

1 0.02
0.03 0.03
1 0.10
0.952423 0.042419
							

2495
2454
							

思路：填充法，找出每个格子的中心点，看它到每个圆心的距离是不是不超过R是就涂色，最后没涂色的就是答案

#include<iostream>
#include<cstring>
#include<cmath>
#include<cstdio>
using namespace std;
const int mm=55;
const double ex=1e-10;
int n;
double r,x,y;
double z=0.02;
bool vis[mm][mm];
double dis(double x1,double y1,double x2,double y2)
{
  double d=(x2-x1)*(x2-x1)+(y2-y1)*(y2-y1);
  d=sqrt(d);
  return d;
}
inline double aabs(double x)
{
  if(x<0)return -x;
  return x;
}
void judge(int a,int b)
{
  double x1=(double)a*z+0.01;
  double y1=(double)b*z+0.01;
  double d=dis(x1,y1,x,y);
  if(d<r)vis[a][b]=1;
  else if(aabs(d-r)<ex)
    vis[a][b]=1;
}

int main()
{
  while(cin>>n>>r)
  { memset(vis,0,sizeof(vis));
    for(int i=0;i<n;++i)
    {
      cin>>x>>y;
      for(int i=0;i<50;i++)
        for(int j=0;j<50;j++)
        judge(i,j);

    }  int ans=0;
        for(int i=0;i<50;i++)
        for(int j=0;j<50;j++)
          if(vis[i][j])
            ++ans;
      ans=2500-ans;
    cout<<ans<<"\n";
  }
}