为连接组件标签/ Blob提取实现LSL

我试图从https://www.lri.fr/~lacas/Publications/JRTIP10.pdf实现Light Speed标签算法。我试图尽可能地遵循本文中描述的算法（第9-12页），但是在等价阶段之后，输出没有意义。为连接组件标签/ Blob提取实现LSL

任何人都有什么想法是什么问题？

void segment(const unsigned * Xi, const unsigned& N, unsigned * ERi, unsigned *RLCi, unsigned& ner) 
{ 
unsigned x1  = 0; 
unsigned f  = 0; 
unsigned er  = 0;  

for (unsigned j = 0; j < N; ++j) 
{ 
    const unsigned x0 = Xi[j]; 
    f = x0^x1; 
    RLCi[er] = j;   
    er = er + f; 
    ERi[j] = er;   
    x1 = x0;   
}  
if (x1 != 0) 
{ 
    RLCi[er] = N; 
} 
er = er + x1; 
ner = er; 
} 

void equivalance(const unsigned& ner, unsigned * RLCi, unsigned * EQ, unsigned * ER0, unsigned * ERA0, unsigned * ERA1, unsigned& nea, const unsigned& N = 0) 
{ 
for (unsigned er = 1; er <= ner; er += 2) 
{   
    int j0 = RLCi[er - 1]; 
    int j1 = RLCi[er] - 1; 
    // Unnecessary given optimization and need for 4-connectivity: 
    // if (j0 > 0) j0 = j0 - 1; 
    // if (j1 < N - 1) j1 = j1 + 1; 

    int er0 = ER0[j0]; 
    int er1 = ER0[j1]; 

    if (!(er0 & 1)) er0 = er0 + 1; 
    if (!(er1 & 1)) er1 = er1 - 1;  

    if (er1 >= er0) // adjacent label 
    {    
     unsigned ea = ERA0[er0]; 
     unsigned a = EQ[ea]; 
     for (unsigned erk = er0 + 2; erk <= er1; ++erk) 
     { 
      unsigned eak = ERA0[erk];    
      unsigned ak = EQ[eak]; 
      if (a < ak) 
      { 
       EQ[eak] = a; 
      } 
      else 
      { 
       a = ak;      
       EQ[ea] = a; 
       ea = eak; 
      }     
     }   
     ERA1[er] = a; 
    } 
    else 
    { 
     nea = nea + 1; 
     ERA1[er] = nea; 
    } 
} 
} 

typedef std::vector<unsigned> value_type; 
void bwlabel(const double* X, unsigned * EA, const unsigned& N, const unsigned& M) 
{ 
unsigned nea = 0; 
const unsigned size = N * M; 

value_type EQ(size, 0), ER(size, 0), ERA(size, 0), A(size, 0), RLC(M * (2 * N), 0), IN(X, X + size), NER(M, 0); 

// Step 1 
for (int m = 0; m < M; ++m) 
{   
    segment(&IN[0] + N * m, N, &ER[0] + N * m, &RLC[0] + m * (2 * N), NER[m]);  
}  
// Step 2 
for (int m = 1; m < M; ++m) 
{   
    equivalance(NER[m], &RLC[0] + m * (2 * N), &EQ[0], &ER[0] + (m - 1) * N, &ERA[0] + (m - 1) * N, &ERA[0] + m * N, nea, N);    
} 
// Step 3 
for (int j = 0; j < size; ++j) 
{ 
    EA[j] = ERA[ER[j]];   
} 

// Step 4 
unsigned na = 0; 
for (int e = 0; e < size; ++e) 
{ 
    if (EQ[e] != e) 
    { 
     A[e] = EQ[EQ[e]]; 
    } 
    else 
    { 
     na = na + 1; 
     A[e] = na; 
    } 
} 
// Step 5 
for (int j = 0; j < size; ++j) 
{ 
    EA[j] = A[EA[j]]; 
} 
} 
IN= 
1 1 0 1 1 0 0 1 1 1 
0 1 1 0 1 0 0 1 1 1 
1 0 1 1 1 1 1 0 1 0 
1 0 0 0 0 1 1 0 1 0 
0 0 1 1 0 0 0 1 1 1 
0 1 1 0 0 0 1 0 1 0 
1 0 1 1 1 1 1 0 0 0 
1 0 1 0 1 0 0 0 1 0 
0 1 1 1 1 0 1 1 0 1 
0 0 1 1 1 0 1 0 0 0 


RLC= 
[0]0 [1]2 [2]3 [3]5 [4]7 [5]10 [6]0 [7]0 [8]0 [9]0 
[0]0 [1]0 [2]0 [3]0 [4]0 [5]0 [6]0 [7]0 [8]0 [9]0 
[0]1 [1]3 [2]4 [3]5 [4]7 [5]10 [6]0 [7]0 [8]0 [9]0 
[0]0 [1]0 [2]0 [3]0 [4]0 [5]0 [6]0 [7]0 [8]0 [9]0 
[0]0 [1]1 [2]2 [3]7 [4]8 [5]9 [6]0 [7]0 [8]0 [9]0 
[0]0 [1]0 [2]0 [3]0 [4]0 [5]0 [6]0 [7]0 [8]0 [9]0 
[0]0 [1]1 [2]5 [3]7 [4]8 [5]9 [6]0 [7]0 [8]0 [9]0 
[0]0 [1]0 [2]0 [3]0 [4]0 [5]0 [6]0 [7]0 [8]0 [9]0 
[0]2 [1]4 [2]7 [3]10 [4]0 [5]0 [6]0 [7]0 [8]0 [9]0 
[0]0 [1]0 [2]0 [3]0 [4]0 [5]0 [6]0 [7]0 [8]0 [9]0 


ER= 
1 1 2 3 3 4 4 5 5 5 
0 1 1 2 3 4 4 5 5 5 
1 2 3 3 3 3 3 4 5 6 
1 2 2 2 2 3 3 4 5 6 
0 0 1 1 2 2 2 3 3 3 
0 1 1 2 2 2 3 4 5 6 
1 2 3 3 3 3 3 4 4 4 
1 2 3 4 5 6 6 6 7 8 
0 1 1 1 1 2 3 3 4 5 
0 0 1 1 1 2 3 4 4 4 


ERA= 
0 0 0 0 0 0 0 0 0 0 
0 0 0 0 0 0 0 0 0 0 
0 0 0 0 0 0 0 0 0 0 
0 0 0 0 0 0 0 0 0 0 
0 1 0 0 0 0 0 0 0 0 
0 0 0 2 0 0 0 0 0 0 
0 0 0 0 0 0 0 0 0 0 
0 0 0 0 0 0 0 3 0 0 
0 0 0 4 0 5 0 0 0 0 
0 0 0 0 0 0 0 0 0 0 


EQ= 
0 0 0 0 0 0 0 0 0 0 
0 0 0 0 0 0 0 0 0 0 
0 0 0 0 0 0 0 0 0 0 
0 0 0 0 0 0 0 0 0 0 
0 0 0 0 0 0 0 0 0 0 
0 0 0 0 0 0 0 0 0 0 
0 0 0 0 0 0 0 0 0 0 
0 0 0 0 0 0 0 0 0 0 
0 0 0 0 0 0 0 0 0 0 
0 0 0 0 0 0 0 0 0 0 


EA= 
1 1 1 1 1 1 1 1 1 1 
1 1 1 1 1 1 1 1 1 1 
1 1 1 1 1 1 1 1 1 1 
1 1 1 1 1 1 1 1 1 1 
1 1 1 1 1 1 1 1 1 1 
1 1 1 1 1 1 1 1 1 1 
1 1 1 1 1 1 1 1 1 1 
1 1 1 1 1 1 1 1 1 1 
1 1 1 1 1 1 1 1 1 1 
1 1 1 1 1 1 1 1 1 1

来源

2015-06-14 Dawud

对于初学者来说，你已经忽略了右边框的补偿变量b。这会影响RLC的结果。

你获得（第一行）：

[0]0 [1]2 [2]3 [3]5 [4]7 [5]10 [6]0 [7]0 [8]0 [9]0

时，你应该获得：

[0]0 [1]1 [2]3 [3]4 [4]7 [5]10 [6]0 [7]0 [8]0 [9]0

感谢。

来源

2016-06-22 08:43:43 Alberto

为连接组件标签/ Blob提取实现LSL

回答

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