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Kind of a Blur

Image blurring occurs when the object being captured is out of the camera‘s focus. The top two figures on the right are an example of an image and its blurred version. Restoring the original image given only the blurred version is one of the most interesting topics in image processing. This process is called deblurring, which will be your task for this problem.
In this problem, all images are in grey-scale (no colours). Images are represented as a 2 dimensional matrix of real numbers, where each cell corresponds to the brightness of the corresponding pixel. Although not mathematically accurate, one way to describe a blurred image is through averaging all the pixels that are within (less than or equal to) a certain Manhattan distance?from each pixel (including the pixel itself ). Here‘s an example of how to calculate the blurring of a 3x3 image with a blurring distance of 1:


Given the blurred version of an image, we are interested in reconstructing the original version assuming that the image was blurred as explained above.

2 2 1
1 1
1 1

3 3 1
19 14 20
12 15 18
13 14 16

4 4 2
14 15 14 15
14 15 14 15
14 15 14 15
14 15 14 15

0 0 0

    1.00    1.00
    1.00    1.00

    2.00   30.00   17.00
   25.00    7.00   13.00
   14.00    0.00   35.00

    1.00   27.00    2.00   28.00
   21.00   12.00   17.00    8.00
   21.00   12.00   17.00    8.00
    1.00   27.00    2.00   28.00
Hint
The Manhattan Distance (sometimes called the Taxicab distance) between
two points is the sum of the (absolute) difference of their coordinates. 
The grid on the lower right illustrates the Manhattan distances from the grayed cell.
 

题解及代码：

     这是一个double精度的高斯消元，方程建立直接暴力枚举每个点与当前点的距离就可以了，注意一下题目中给出的输入 先是列数，后是行数T^T

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
using namespace std;
const double eps=0.00000001;

double a[110][110];
double x[110];
int equ,var;

void Debug()
{
    for(int i=0;i<equ;i++)
    {
        for(int j=0;j<=var;j++)
        printf("%.2lf ",a[i][j]);
        puts("");
    }
    puts("");
}



void Gauss()
{
    int k,col,max_r;

    for(k=0,col=0;k<equ&&col<var;k++,col++)
    {
       max_r=k;
       for(int i=k+1;i<equ;i++)
       if(fabs(a[i][col])-fabs(a[max_r][col])>eps) max_r=i;

       if(max_r!=k)
       for(int i=0;i<=var;i++)
       {
          swap(a[max_r][i],a[k][i]);
       }

       if(fabs(a[k][col])<eps)
       {
          k--;
          continue;
       }


       for(int i=k+1;i<equ;i++)
       {
          if(fabs(a[i][col])>eps)
          {
             double t=a[i][col]/a[k][col];
             for(int j=col;j<=var;j++)
             {
                a[i][j]=a[i][j]-a[k][j]*t;
             }
          }
       }

    }
     //Debug();

    for(int j=var-1;j>=0;j--)
    {
        if(fabs(a[j][j])<eps) continue;
        double temp=a[j][var];
        for(int r=j+1;r<var;r++)
        temp=temp-a[j][r]*x[r];

        x[j]=temp/a[j][j];
    }

}

void init(int m,int n,int d)               //系数矩阵初始化
{
    var=equ=m*n;
    int D,B,L,R;
    memset(a,0,sizeof(a));
    memset(x,0,sizeof(x));
    int pos=0;
    double k=0;
    for(int i=0;i<m;i++)
    for(int j=0;j<n;j++)
    {
        k=0;
        pos=i*n+j;
        for(int l=0;l<m;l++)
        for(int r=0;r<n;r++)
        if(abs(l-i)+abs(r-j)<=d)
        {
            k+=1.0;
            a[pos][l*n+r]=1.0;
        }
        a[pos][var]=k;
    }
    //Debug();
}

int main()
{
  int n,m,d;
  double X;
  bool fis=true;
  while(scanf("%d%d%d",&n,&m,&d))
  {
      if(n==0&&m==0&&d==0) break;
      if(!fis) puts("");
      else fis=false;

      init(m,n,d);
      for(int i=0;i<equ;i++)
      {
          scanf("%lf",&X);
          a[i][var]*=X;
      }
      //Debug();
      Gauss();
      for(int i=0;i<equ;i++)
      {
          printf("%8.2lf",x[i]);
          if((i+1)%n==0) printf("\n");
      }
  }
  return 0;
}

hdu 3359 Kind of a Blur

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原文地址：http://blog.csdn.net/knight_kaka/article/details/39159085