encuentra el algoritmo de submatriz más grande

Question

Tengo una matriz N * N (N = 2 a 10000) de números que pueden oscilar entre 0 y 1000. ¿Cómo puedo encontrar la submatriz más grande (rectangular) que consta del mismo número?

Ejemplo:

 1 2 3 4 5 
    -- -- -- -- -- 
1 | 10 9 9 9 80 
2 | 5 9 9 9 10 
3 | 85 86 54 45 45 
4 | 15 21 5 1 0 
5 | 5 6 88 11 10 

La salida debería ser el área de la submatriz, seguido de coordenadas basadas en 1 de su parte superior elemento izquierdo. Para el ejemplo, sería (6, 2, 1) porque hay seis 9 s situados en la columna 2, fila 1.

Answer 1

Este es un trabajo en progreso

pensé en este problema y creo que puede tener una O(w*h) algoritmo.

La idea es la siguiente:

• para cualquier (i,j) calcular el número más alto de células con el mismo valor en la columna j partir de (i,j). Almacene estos valores como heights[i][j].
• crear un vector vacío de la matriz sub (a LIFO)
• para todos consecutivas: i
  • para todos columna: j
    • pop todo matriz sub cuya height > heights[i][j]. Debido a que la submatriz con altura>heights[i][j] no puede continuar en esta celda
    • empuje una submatriz definido por (i,j,heights[i][j]) donde j es el más alejado de coordenadas en el que podemos ajustar una submatriz de altura: heights[i][j]
    • actualización de la sub matriz máx

La parte engañosa está en el bucle interno. Utilizo algo similar al algoritmo de subventana max para asegurar que sea O(1) en promedio para cada celda.

Trataré de formular una prueba, pero mientras tanto aquí está el código.

#include <algorithm> 
#include <iterator> 
#include <iostream> 
#include <ostream> 
#include <vector> 

typedef std::vector<int> row_t; 
typedef std::vector<row_t> matrix_t; 

std::size_t height(matrix_t const& M) { return M.size(); } 
std::size_t width (matrix_t const& M) { return M.size() ? M[0].size() : 0u; } 

std::ostream& operator<<(std::ostream& out, matrix_t const& M) { 
    for(unsigned i=0; i<height(M); ++i) { 
    std::copy(begin(M[i]), end(M[i]), 
      std::ostream_iterator<int>(out, ", ")); 
    out << std::endl; 
    } 
    return out; 
} 

struct sub_matrix_t { 
    int i, j, h, w; 
    sub_matrix_t(): i(0),j(0),h(0),w(1) {} 
    sub_matrix_t(int i_,int j_,int h_,int w_):i(i_),j(j_),h(h_),w(w_) {} 
    bool operator<(sub_matrix_t const& rhs) const { return (w*h)<(rhs.w*rhs.h); } 
}; 


// Pop all sub_matrix from the vector keeping only those with an height 
// inferior to the passed height. 
// Compute the max sub matrix while removing sub matrix with height > h 
void pop_sub_m(std::vector<sub_matrix_t>& subs, 
      int i, int j, int h, sub_matrix_t& max_m) { 

    sub_matrix_t sub_m(i, j, h, 1); 

    while(subs.size() && subs.back().h >= h) { 
    sub_m = subs.back(); 
    subs.pop_back(); 
    sub_m.w = j-sub_m.j; 
    max_m = std::max(max_m, sub_m); 
    } 

    // Now sub_m.{i,j} is updated to the farest coordinates where there is a 
    // submatrix with heights >= h 

    // If we don't cut the current height (because we changed value) update 
    // the current max submatrix 
    if(h > 0) { 
    sub_m.h = h; 
    sub_m.w = j-sub_m.j+1; 
    max_m = std::max(max_m, sub_m); 
    subs.push_back(sub_m); 
    } 
} 

void push_sub_m(std::vector<sub_matrix_t>& subs, 
     int i, int j, int h, sub_matrix_t& max_m) { 
    if(subs.empty() || subs.back().h < h) 
    subs.emplace_back(i, j, h, 1); 
} 

void solve(matrix_t const& M, sub_matrix_t& max_m) { 
    // Initialize answer suitable for an empty matrix 
    max_m = sub_matrix_t(); 
    if(height(M) == 0 || width(M) == 0) return; 

    // 1) Compute the heights of columns of the same values 
    matrix_t heights(height(M), row_t(width(M), 1)); 
    for(unsigned i=height(M)-1; i>0; --i) 
    for(unsigned j=0; j<width(M); ++j) 
     if(M[i-1][j]==M[i][j]) 
    heights[i-1][j] = heights[i][j]+1; 

    // 2) Run through all columns heights to compute local sub matrices 
    std::vector<sub_matrix_t> subs; 
    for(int i=height(M)-1; i>=0; --i) { 
    push_sub_m(subs, i, 0, heights[i][0], max_m); 
    for(unsigned j=1; j<width(M); ++j) { 
     bool same_val = (M[i][j]==M[i][j-1]); 
     int pop_height = (same_val) ? heights[i][j] : 0; 
     int pop_j  = (same_val) ? j    : j-1; 
     pop_sub_m (subs, i, pop_j, pop_height, max_m); 
     push_sub_m(subs, i, j,  heights[i][j], max_m); 
    } 
    pop_sub_m(subs, i, width(M)-1, 0, max_m); 
    } 
} 

matrix_t M1{ 
    {10, 9, 9, 9, 80}, 
    { 5, 9, 9, 9, 10}, 
    {85, 86, 54, 45, 45}, 
    {15, 21, 5, 1, 0}, 
    { 5, 6, 88, 11, 10}, 
}; 

matrix_t M2{ 
    {10, 19, 9, 29, 80}, 
    { 5, 9, 9, 9, 10}, 
    { 9, 9, 54, 45, 45}, 
    { 9, 9, 5, 1, 0}, 
    { 5, 6, 88, 11, 10}, 
}; 


int main() { 
    sub_matrix_t answer; 

    std::cout << M1 << std::endl; 
    solve(M1, answer); 
    std::cout << '(' << (answer.w*answer.h) 
     << ',' << (answer.j+1) << ',' << (answer.i+1) << ')' 
     << std::endl; 

    answer = sub_matrix_t(); 
    std::cout << M2 << std::endl; 
    solve(M2, answer); 
    std::cout << '(' << (answer.w*answer.h) 
     << ',' << (answer.j+1) << ',' << (answer.i+1) << ')' 
     << std::endl; 
} 

Answer 2

Ésta es una Filas del pedido * Columnas Solución

Funciona

• a partir de la parte inferior de la matriz, y determinar cuántos elementos hay debajo de cada partido número en una columna. Esto se hace en tiempo O (MN) (muy trivial)
• Luego va de arriba a abajo & de izquierda a derecha y ve si un número determinado coincide con el número de la izquierda. Si es así, realiza un seguimiento de cómo las alturas se relacionan entre sí para rastrear las posibles formas de rectángulo

Aquí está una implementación de trabajo de python.Disculpas ya que no estoy seguro de cómo obtener el resaltado de sintaxis de trabajo

# this program finds the largest area in an array where all the elements have the same value 
# It solves in O(rows * columns) time using O(rows*columns) space using dynamic programming 




def max_area_subarray(array): 

    rows = len(array) 
    if (rows == 0): 
     return [[]] 
    columns = len(array[0]) 


    # initialize a blank new array 
    # this will hold max elements of the same value in a column 
    new_array = [] 
    for i in range(0,rows-1): 
     new_array.append([0] * columns) 

    # start with the bottom row, these all of 1 element of the same type 
    # below them, including themselves 
    new_array.append([1] * columns) 

    # go from the second to bottom row up, finding how many contiguous 
    # elements of the same type there are 
    for i in range(rows-2,-1,-1): 
     for j in range(columns-1,-1,-1): 
      if (array[i][j] == array[i+1][j]): 
       new_array[i][j] = new_array[i+1][j]+1 
      else: 
       new_array[i][j] = 1 


    # go left to right and match up the max areas 
    max_area = 0 
    top = 0 
    bottom = 0 
    left = 0 
    right = 0 
    for i in range(0,rows): 
     running_height =[[0,0,0]] 
     for j in range(0,columns): 

      matched = False 
      if (j > 0): # if this isn't the leftmost column 
       if (array[i][j] == array[i][j-1]): 
        # this matches the array to the left 
        # keep track of if this is a longer column, shorter column, or same as 
        # the one on the left 
        matched = True 

        while(new_array[i][j] < running_height[-1][0]): 
         # this is less than the one on the left, pop that running 
         # height from the list, and add it's columns to the smaller 
         # running height below it 
         if (running_height[-1][1] > max_area): 
          max_area = running_height[-1][1] 
          top = i 
          right = j-1 
          bottom = i + running_height[-1][0]-1 
          left = j - running_height[-1][2] 

         previous_column = running_height.pop() 
         num_columns = previous_column[2] 

         if (len(running_height) > 0): 
          running_height[-1][1] += running_height[-1][0] * (num_columns) 
          running_height[-1][2] += num_columns 

         else: 
          # for instance, if we have heights 2,2,1 
          # this will trigger on the 1 after we pop the 2 out, and save the current 
          # height of 1, the running area of 3, and running columsn of 3 
          running_height.append([new_array[i][j],new_array[i][j]*(num_columns),num_columns]) 


        if (new_array[i][j] > running_height[-1][0]): 
         # longer then the one on the left 
         # append this height and area 
         running_height.append([new_array[i][j],new_array[i][j],1]) 
        elif (new_array[i][j] == running_height[-1][0]): 
         # same as the one on the left, add this area to the one on the left 
         running_height[-1][1] += new_array[i][j] 
         running_height[-1][2] += 1 



      if (matched == False or j == columns -1): 
       while(running_height): 
        # unwind the maximums & see if this is the new max area 
        if (running_height[-1][1] > max_area): 
         max_area = running_height[-1][1] 
         top = i 
         right = j 
         bottom = i + running_height[-1][0]-1 
         left = j - running_height[-1][2]+1 

         # this wasn't a match, so move everything one bay to the left 
         if (matched== False): 
          right = right-1 
          left = left-1 


        previous_column = running_height.pop() 
        num_columns = previous_column[2] 
        if (len(running_height) > 0): 
         running_height[-1][1] += running_height[-1][0] * num_columns 
         running_height[-1][2] += num_columns 

      if (matched == False): 
       # this is either the left column, or we don't match to the column to the left, so reset 
       running_height = [[new_array[i][j],new_array[i][j],1]] 
       if (running_height[-1][1] > max_area): 
        max_area = running_height[-1][1] 
        top = i 
        right = j 
        bottom = i + running_height[-1][0]-1 
        left = j - running_height[-1][2]+1 


    max_array = [] 
    for i in range(top,bottom+1): 
     max_array.append(array[i][left:right+1]) 


    return max_array 



numbers = [[6,4,1,9],[5,2,2,7],[2,2,2,1],[2,3,1,5]] 

for row in numbers: 
    print row 

print 
print 

max_array = max_area_subarray(numbers)  


max_area = len(max_array) * len(max_array[0]) 
print 'max area is ',max_area 
print 
for row in max_array: 
    print row