算法设计：如何计算全为1的最大矩形二进制子矩阵？

2021年3月20日16:54:07 发表评论 670 次浏览

本文概述

给定一个二进制矩阵, 找到全为1的最大尺寸的矩形二进制子矩阵。

例子：

Input:
0 1 1 0
1 1 1 1
1 1 1 1
1 1 0 0
Output :
1 1 1 1
1 1 1 1
Explanation : 
The largest rectangle with only 1's is from 
(1, 0) to (2, 3) which is
1 1 1 1
1 1 1 1 

Input:
0 1 1
1 1 1
0 1 1      
Output:
1 1
1 1
1 1
Explanation : 
The largest rectangle with only 1's is from 
(0, 1) to (2, 2) which is
1 1
1 1
1 1

推荐：请在"实践首先, 在继续解决方案之前。

已经有讨论过的算法了基于动态编程的解决方案, 以寻找最大的平方与1.

方法：

在这篇文章中, 讨论了一种有趣的方法, 该方法使用

直方图下的最大矩形

作为子例程。

如果给出直方图的条形高度, 则可以找到直方图的最大面积。这样, 在每一行中, 可以找到直方图的最大条形区域。要使最大的矩形充满1, 请在上一行的基础上更新下一行, 并在直方图下找到最大的面积, 即, 将每个1填充为正方形, 将0填充为一个空正方形, 并将每行作为底数。

插图：

Input :
0 1 1 0
1 1 1 1
1 1 1 1
1 1 0 0
Step 1: 
    0 1 1 0  maximum area  = 2
Step 2:
    row 1  1 2 2 1  area = 4, maximum area becomes 4
    row 2  2 3 3 2  area = 8, maximum area becomes 8
    row 3  3 4 0 0  area = 6, maximum area remains 8

算法：

  1. 运行循环以遍历各行。
  2. 现在, 如果当前行不是第一行, 则按如下方式更新该行, 如果matrix [i] [j]不为零, 则matrix [i] [j] = matrix [i-1] [j] + matrix [i ] [j]。
  3. 找到直方图下方的最大矩形区域, 将第i行视为直方图的条形高度。可以按照本文给出的方法进行计算直方图中最大的矩形区域
  4. 对所有行执行前两个步骤, 并打印所有行的最大面积。

注意

：强烈建议参考

这个

首先发布, 因为大多数代码是从那里获取的。

实现

C ++

// C++ program to find largest rectangle with all 1s
// in a binary matrix
#include <bits/stdc++.h>
using namespace std;
  
// Rows and columns in input matrix
#define R 4
#define C 4
  
// Finds the maximum area under the histogram represented
// by histogram.  See below article for details.
// https:// www.lsbin.org/largest-rectangle-under-histogram/
int maxHist( int row[])
{
     // Create an empty stack. The stack holds indexes of
     // hist[] array/ The bars stored in stack are always
     // in increasing order of their heights.
     stack< int > result;
  
     int top_val; // Top of stack
  
     int max_area = 0; // Initialize max area in current
     // row (or histogram)
  
     int area = 0; // Initialize area with current top
  
     // Run through all bars of given histogram (or row)
     int i = 0;
     while (i < C) {
         // If this bar is higher than the bar on top stack, // push it to stack
         if (result.empty() || row[result.top()] <= row[i])
             result.push(i++);
  
         else {
             // If this bar is lower than top of stack, then
             // calculate area of rectangle with stack top as
             // the smallest (or minimum height) bar. 'i' is
             // 'right index' for the top and element before
             // top in stack is 'left index'
             top_val = row[result.top()];
             result.pop();
             area = top_val * i;
  
             if (!result.empty())
                 area = top_val * (i - result.top() - 1);
             max_area = max(area, max_area);
         }
     }
  
     // Now pop the remaining bars from stack and calculate area
     // with every popped bar as the smallest bar
     while (!result.empty()) {
         top_val = row[result.top()];
         result.pop();
         area = top_val * i;
         if (!result.empty())
             area = top_val * (i - result.top() - 1);
  
         max_area = max(area, max_area);
     }
     return max_area;
}
  
// Returns area of the largest rectangle with all 1s in A[][]
int maxRectangle( int A[][C])
{
     // Calculate area for first row and initialize it as
     // result
     int result = maxHist(A[0]);
  
     // iterate over row to find maximum rectangular area
     // considering each row as histogram
     for ( int i = 1; i < R; i++) {
  
         for ( int j = 0; j < C; j++)
  
             // if A[i][j] is 1 then add A[i -1][j]
             if (A[i][j])
                 A[i][j] += A[i - 1][j];
  
         // Update result if area with current row (as last row)
         // of rectangle) is more
         result = max(result, maxHist(A[i]));
     }
  
     return result;
}
  
// Driver code
int main()
{
     int A[][C] = {
         { 0, 1, 1, 0 }, { 1, 1, 1, 1 }, { 1, 1, 1, 1 }, { 1, 1, 0, 0 }, };
  
     cout << "Area of maximum rectangle is "
          << maxRectangle(A);
  
     return 0;
}

Java

// Java program to find largest rectangle with all 1s
// in a binary matrix
import java.io.*;
import java.util.*;
  
class GFG {
     // Finds the maximum area under the histogram represented
     // by histogram.  See below article for details.
     // https:// www.lsbin.org/largest-rectangle-under-histogram/
     static int maxHist( int R, int C, int row[])
     {
         // Create an empty stack. The stack holds indexes of
         // hist[] array/ The bars stored in stack are always
         // in increasing order of their heights.
         Stack<Integer> result = new Stack<Integer>();
  
         int top_val; // Top of stack
  
         int max_area = 0 ; // Initialize max area in current
         // row (or histogram)
  
         int area = 0 ; // Initialize area with current top
  
         // Run through all bars of given histogram (or row)
         int i = 0 ;
         while (i < C) {
             // If this bar is higher than the bar on top stack, // push it to stack
             if (result.empty() || row[result.peek()] <= row[i])
                 result.push(i++);
  
             else {
                 // If this bar is lower than top of stack, then
                 // calculate area of rectangle with stack top as
                 // the smallest (or minimum height) bar. 'i' is
                 // 'right index' for the top and element before
                 // top in stack is 'left index'
                 top_val = row[result.peek()];
                 result.pop();
                 area = top_val * i;
  
                 if (!result.empty())
                     area = top_val * (i - result.peek() - 1 );
                 max_area = Math.max(area, max_area);
             }
         }
  
         // Now pop the remaining bars from stack and calculate
         // area with every popped bar as the smallest bar
         while (!result.empty()) {
             top_val = row[result.peek()];
             result.pop();
             area = top_val * i;
             if (!result.empty())
                 area = top_val * (i - result.peek() - 1 );
  
             max_area = Math.max(area, max_area);
         }
         return max_area;
     }
  
     // Returns area of the largest rectangle with all 1s in
     // A[][]
     static int maxRectangle( int R, int C, int A[][])
     {
         // Calculate area for first row and initialize it as
         // result
         int result = maxHist(R, C, A[ 0 ]);
  
         // iterate over row to find maximum rectangular area
         // considering each row as histogram
         for ( int i = 1 ; i < R; i++) {
  
             for ( int j = 0 ; j < C; j++)
  
                 // if A[i][j] is 1 then add A[i -1][j]
                 if (A[i][j] == 1 )
                     A[i][j] += A[i - 1 ][j];
  
             // Update result if area with current row (as last
             // row of rectangle) is more
             result = Math.max(result, maxHist(R, C, A[i]));
         }
  
         return result;
     }
  
     // Driver code
     public static void main(String[] args)
     {
         int R = 4 ;
         int C = 4 ;
  
         int A[][] = {
             { 0 , 1 , 1 , 0 }, { 1 , 1 , 1 , 1 }, { 1 , 1 , 1 , 1 }, { 1 , 1 , 0 , 0 }, };
         System.out.print( "Area of maximum rectangle is " + maxRectangle(R, C, A));
     }
}
  
// Contributed by Prakriti Gupta

Python3

# Python3 program to find largest rectangle 
# with all 1s in a binary matrix 
  
# Rows and columns in input matrix 
R = 4
C = 4
  
# Finds the maximum area under the histogram represented 
# by histogram. See below article for details. 
# https:# www.lsbin.org / largest-rectangle-under-histogram / def maxHist(row):
  
     # Create an empty stack. The stack holds 
     # indexes of hist array / The bars stored  
     # in stack are always in increasing order 
     # of their heights. 
     result = []
  
     top_val = 0     # Top of stack 
  
     max_area = 0 # Initialize max area in current 
                  # row (or histogram) 
  
     area = 0 # Initialize area with current top 
  
     # Run through all bars of given
     # histogram (or row) 
     i = 0
     while (i < C): 
      
         # If this bar is higher than the 
         # bar on top stack, push it to stack 
         if ( len (result) = = 0 ) or (row[result[ 0 ]] < = row[i]):
             result.append(i)
             i + = 1
         else :
          
             # If this bar is lower than top of stack, # then calculate area of rectangle with 
             # stack top as the smallest (or minimum 
             # height) bar. 'i' is 'right index' for 
             # the top and element before top in stack
             # is 'left index' 
             top_val = row[result[ 0 ]] 
             result.pop( 0 ) 
             area = top_val * i 
  
             if ( len (result)):
                 area = top_val * (i - result[ 0 ] - 1 ) 
             max_area = max (area, max_area) 
          
     # Now pop the remaining bars from stack 
     # and calculate area with every popped
     # bar as the smallest bar 
     while ( len (result)):
         top_val = row[result[ 0 ]] 
         result.pop( 0 ) 
         area = top_val * i 
         if ( len (result)): 
             area = top_val * (i - result[ 0 ] - 1 ) 
  
         max_area = max (area, max_area) 
      
     return max_area 
  
# Returns area of the largest rectangle 
# with all 1s in A 
def maxRectangle(A):
      
     # Calculate area for first row and 
     # initialize it as result 
     result = maxHist(A[ 0 ]) 
  
     # iterate over row to find maximum rectangular 
     # area considering each row as histogram 
     for i in range ( 1 , R):
      
         for j in range (C):
  
             # if A[i][j] is 1 then add A[i -1][j] 
             if (A[i][j]):
                 A[i][j] + = A[i - 1 ][j] 
  
         # Update result if area with current 
         # row (as last row) of rectangle) is more 
         result = max (result, maxHist(A[i])) 
      
     return result 
      
# Driver Code 
if __name__ = = '__main__' :
     A = [[ 0 , 1 , 1 , 0 ], [ 1 , 1 , 1 , 1 ], [ 1 , 1 , 1 , 1 ], [ 1 , 1 , 0 , 0 ]] 
  
     print ( "Area of maximum rectangle is" , maxRectangle(A))
      
# This code is contributed 
# by SHUBHAMSINGH10

C#

// C# program to find largest rectangle
// with all 1s in a binary matrix
using System;
using System.Collections.Generic;
  
class GFG {
     // Finds the maximum area under the
     // histogram represented by histogram.
     // See below article for details.
     // https:// www.lsbin.org/largest-rectangle-under-histogram/
     public static int maxHist( int R, int C, int [] row)
     {
         // Create an empty stack. The stack
         // holds indexes of hist[] array.
         // The bars stored in stack are always
         // in increasing order of their heights.
         Stack< int > result = new Stack< int >();
  
         int top_val; // Top of stack
  
         int max_area = 0; // Initialize max area in
         // current row (or histogram)
  
         int area = 0; // Initialize area with
         // current top
  
         // Run through all bars of
         // given histogram (or row)
         int i = 0;
         while (i < C) {
             // If this bar is higher than the
             // bar on top stack, push it to stack
             if (result.Count == 0 || row[result.Peek()] <= row[i]) {
                 result.Push(i++);
             }
  
             else {
                 // If this bar is lower than top
                 // of stack, then calculate area of
                 // rectangle with stack top as
                 // the smallest (or minimum height)
                 // bar. 'i' is 'right index' for
                 // the top and element before
                 // top in stack is 'left index'
                 top_val = row[result.Peek()];
                 result.Pop();
                 area = top_val * i;
  
                 if (result.Count > 0) {
                     area = top_val * (i - result.Peek() - 1);
                 }
                 max_area = Math.Max(area, max_area);
             }
         }
  
         // Now pop the remaining bars from
         // stack and calculate area with
         // every popped bar as the smallest bar
         while (result.Count > 0) {
             top_val = row[result.Peek()];
             result.Pop();
             area = top_val * i;
             if (result.Count > 0) {
                 area = top_val * (i - result.Peek() - 1);
             }
  
             max_area = Math.Max(area, max_area);
         }
         return max_area;
     }
  
     // Returns area of the largest
     // rectangle with all 1s in A[][]
     public static int maxRectangle( int R, int C, int [][] A)
     {
         // Calculate area for first row
         // and initialize it as result
         int result = maxHist(R, C, A[0]);
  
         // iterate over row to find
         // maximum rectangular area
         // considering each row as histogram
         for ( int i = 1; i < R; i++) {
             for ( int j = 0; j < C; j++) {
  
                 // if A[i][j] is 1 then
                 // add A[i -1][j]
                 if (A[i][j] == 1) {
                     A[i][j] += A[i - 1][j];
                 }
             }
  
             // Update result if area with current
             // row (as last row of rectangle) is more
             result = Math.Max(result, maxHist(R, C, A[i]));
         }
  
         return result;
     }
  
     // Driver code
     public static void Main( string [] args)
     {
         int R = 4;
         int C = 4;
  
         int [][] A = new int [][] {
             new int [] { 0, 1, 1, 0 }, new int [] { 1, 1, 1, 1 }, new int [] { 1, 1, 1, 1 }, new int [] { 1, 1, 0, 0 }
         };
         Console.Write( "Area of maximum rectangle is " + maxRectangle(R, C, A));
     }
}
  
// This code is contributed by Shrikant13

输出：

Area of maximum rectangle is 8

复杂度分析：

  • 时间复杂度：O(R x C)。
    矩阵只需要遍历一次, 因此时间复杂度为O(R X C)
  • 空间复杂度：O(C)。
    需要使用堆栈来存储列, 因此空间复杂度为O(C)

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