NEFU 653 Grids(water)

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本文探讨在网格上划分并覆盖单位格子的问题，给定等大小的圆圈数量和半径，计算未被圆圈覆盖的单位格子总数。通过填充法，识别每个单位格子与圆心的距离是否小于等于圆的半径来判断覆盖情况。

Grids

Time Limit 1000ms

Memory Limit 65536K

description

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.

input

　　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).

output

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

sample_input

1 0.02
0.03 0.03
1 0.10
0.952423 0.042419

sample_output

2495
2454

hint

sourc

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";
  }
}