给定一个二进制矩阵，找出全为1的最大大小的平方子矩阵。
例如，考虑下面的二进制矩阵。 
算法：
令给定的二进制矩阵为M [R] [C]。该算法的思想是构造一个辅助大小矩阵S [] []，其中每个条目S [i] [j]表示正方形子矩阵的大小，其中所有1包括M [i] [j]，其中M [ i] [j]是子矩阵中最右边和最下面的条目。
1) Construct a sum matrix S[R][C] for the given M[R][C].
a) Copy first row and first columns as it is from M[][] to S[][]
b) For other entries, use following expressions to construct S[][]
If M[i][j] is 1 then
S[i][j] = min(S[i][j-1], S[i-1][j], S[i-1][j-1]) + 1
Else /*If M[i][j] is 0*/
S[i][j] = 0
2) Find the maximum entry in S[R][C]
3) Using the value and coordinates of maximum entry in S[i], print
sub-matrix of M[][]对于上述示例中给定的M [R] [C]，构造的S [R] [C]将为：
0 1 1 0 1
1 1 0 1 0
0 1 1 1 0
1 1 2 2 0
1 2 2 3 1
0 0 0 0 0上面的矩阵中最大条目的值为3，条目的坐标为(4，3)。使用最大值及其坐标，我们可以找到所需的子矩阵。
C++
// C++ code for Maximum size square
// sub-matrix with all 1s
#include
#define bool int
#define R 6
#define C 5
using namespace std;
void printMaxSubSquare(bool M[R][C])
{
int i,j;
int S[R][C];
int max_of_s, max_i, max_j;
/* Set first column of S[][]*/
for(i = 0; i < R; i++)
S[i][0] = M[i][0];
/* Set first row of S[][]*/
for(j = 0; j < C; j++)
S[0][j] = M[0][j];
/* Construct other entries of S[][]*/
for(i = 1; i < R; i++)
{
for(j = 1; j < C; j++)
{
if(M[i][j] == 1)
S[i][j] = min(S[i][j-1],min( S[i-1][j],
S[i-1][j-1])) + 1;
else
S[i][j] = 0;
}
}
/* Find the maximum entry, and indexes of maximum entry
in S[][] */
max_of_s = S[0][0]; max_i = 0; max_j = 0;
for(i = 0; i < R; i++)
{
for(j = 0; j < C; j++)
{
if(max_of_s < S[i][j])
{
max_of_s = S[i][j];
max_i = i;
max_j = j;
}
}
}
cout<<"Maximum size sub-matrix is: \n";
for(i = max_i; i > max_i - max_of_s; i--)
{
for(j = max_j; j > max_j - max_of_s; j--)
{
cout << M[i][j] << " ";
}
cout << "\n";
}
}
/* Driver code */
int main()
{
bool M[R][C] = {{0, 1, 1, 0, 1},
{1, 1, 0, 1, 0},
{0, 1, 1, 1, 0},
{1, 1, 1, 1, 0},
{1, 1, 1, 1, 1},
{0, 0, 0, 0, 0}};
printMaxSubSquare(M);
}
// This is code is contributed by rathbhupendra C
// C code for Maximum size square
// sub-matrix with all 1s
#include
#define bool int
#define R 6
#define C 5
void printMaxSubSquare(bool M[R][C])
{
int i,j;
int S[R][C];
int max_of_s, max_i, max_j;
/* Set first column of S[][]*/
for(i = 0; i < R; i++)
S[i][0] = M[i][0];
/* Set first row of S[][]*/
for(j = 0; j < C; j++)
S[0][j] = M[0][j];
/* Construct other entries of S[][]*/
for(i = 1; i < R; i++)
{
for(j = 1; j < C; j++)
{
if(M[i][j] == 1)
S[i][j] = min(S[i][j-1], S[i-1][j],
S[i-1][j-1]) + 1;
else
S[i][j] = 0;
}
}
/* Find the maximum entry, and indexes of maximum entry
in S[][] */
max_of_s = S[0][0]; max_i = 0; max_j = 0;
for(i = 0; i < R; i++)
{
for(j = 0; j < C; j++)
{
if(max_of_s < S[i][j])
{
max_of_s = S[i][j];
max_i = i;
max_j = j;
}
}
}
printf("Maximum size sub-matrix is: \n");
for(i = max_i; i > max_i - max_of_s; i--)
{
for(j = max_j; j > max_j - max_of_s; j--)
{
printf("%d ", M[i][j]);
}
printf("\n");
}
}
/* UTILITY FUNCTIONS */
/* Function to get minimum of three values */
int min(int a, int b, int c)
{
int m = a;
if (m > b)
m = b;
if (m > c)
m = c;
return m;
}
/* Driver function to test above functions */
int main()
{
bool M[R][C] = {{0, 1, 1, 0, 1},
{1, 1, 0, 1, 0},
{0, 1, 1, 1, 0},
{1, 1, 1, 1, 0},
{1, 1, 1, 1, 1},
{0, 0, 0, 0, 0}};
printMaxSubSquare(M);
getchar();
} Java
// JAVA Code for Maximum size square
// sub-matrix with all 1s
public class GFG
{
// method for Maximum size square sub-matrix with all 1s
static void printMaxSubSquare(int M[][])
{
int i,j;
int R = M.length; //no of rows in M[][]
int C = M[0].length; //no of columns in M[][]
int S[][] = new int[R][C];
int max_of_s, max_i, max_j;
/* Set first column of S[][]*/
for(i = 0; i < R; i++)
S[i][0] = M[i][0];
/* Set first row of S[][]*/
for(j = 0; j < C; j++)
S[0][j] = M[0][j];
/* Construct other entries of S[][]*/
for(i = 1; i < R; i++)
{
for(j = 1; j < C; j++)
{
if(M[i][j] == 1)
S[i][j] = Math.min(S[i][j-1],
Math.min(S[i-1][j], S[i-1][j-1])) + 1;
else
S[i][j] = 0;
}
}
/* Find the maximum entry, and indexes of maximum entry
in S[][] */
max_of_s = S[0][0]; max_i = 0; max_j = 0;
for(i = 0; i < R; i++)
{
for(j = 0; j < C; j++)
{
if(max_of_s < S[i][j])
{
max_of_s = S[i][j];
max_i = i;
max_j = j;
}
}
}
System.out.println("Maximum size sub-matrix is: ");
for(i = max_i; i > max_i - max_of_s; i--)
{
for(j = max_j; j > max_j - max_of_s; j--)
{
System.out.print(M[i][j] + " ");
}
System.out.println();
}
}
// Driver program
public static void main(String[] args)
{
int M[][] = {{0, 1, 1, 0, 1},
{1, 1, 0, 1, 0},
{0, 1, 1, 1, 0},
{1, 1, 1, 1, 0},
{1, 1, 1, 1, 1},
{0, 0, 0, 0, 0}};
printMaxSubSquare(M);
}
}Python3
# Python3 code for Maximum size
# square sub-matrix with all 1s
def printMaxSubSquare(M):
R = len(M) # no. of rows in M[][]
C = len(M[0]) # no. of columns in M[][]
S = [[0 for k in range(C)] for l in range(R)]
# here we have set the first row and column of S[][]
# Construct other entries
for i in range(1, R):
for j in range(1, C):
if (M[i][j] == 1):
S[i][j] = min(S[i][j-1], S[i-1][j],
S[i-1][j-1]) + 1
else:
S[i][j] = 0
# Find the maximum entry and
# indices of maximum entry in S[][]
max_of_s = S[0][0]
max_i = 0
max_j = 0
for i in range(R):
for j in range(C):
if (max_of_s < S[i][j]):
max_of_s = S[i][j]
max_i = i
max_j = j
print("Maximum size sub-matrix is: ")
for i in range(max_i, max_i - max_of_s, -1):
for j in range(max_j, max_j - max_of_s, -1):
print (M[i][j], end = " ")
print("")
# Driver Program
M = [[0, 1, 1, 0, 1],
[1, 1, 0, 1, 0],
[0, 1, 1, 1, 0],
[1, 1, 1, 1, 0],
[1, 1, 1, 1, 1],
[0, 0, 0, 0, 0]]
printMaxSubSquare(M)
# This code is contributed by Soumen GhoshC#
// C# Code for Maximum size square
// sub-matrix with all 1s
using System;
public class GFG
{
// method for Maximum size square sub-matrix with all 1s
static void printMaxSubSquare(int [,]M)
{
int i,j;
//no of rows in M[,]
int R = M.GetLength(0);
//no of columns in M[,]
int C = M.GetLength(1);
int [,]S = new int[R,C];
int max_of_s, max_i, max_j;
/* Set first column of S[,]*/
for(i = 0; i < R; i++)
S[i,0] = M[i,0];
/* Set first row of S[][]*/
for(j = 0; j < C; j++)
S[0,j] = M[0,j];
/* Construct other entries of S[,]*/
for(i = 1; i < R; i++)
{
for(j = 1; j < C; j++)
{
if(M[i,j] == 1)
S[i,j] = Math.Min(S[i,j-1],
Math.Min(S[i-1,j], S[i-1,j-1])) + 1;
else
S[i,j] = 0;
}
}
/* Find the maximum entry, and indexes of
maximum entry in S[,] */
max_of_s = S[0,0]; max_i = 0; max_j = 0;
for(i = 0; i < R; i++)
{
for(j = 0; j < C; j++)
{
if(max_of_s < S[i,j])
{
max_of_s = S[i,j];
max_i = i;
max_j = j;
}
}
}
Console.WriteLine("Maximum size sub-matrix is: ");
for(i = max_i; i > max_i - max_of_s; i--)
{
for(j = max_j; j > max_j - max_of_s; j--)
{
Console.Write(M[i,j] + " ");
}
Console.WriteLine();
}
}
// Driver program
public static void Main()
{
int [,]M = new int[6,5]{{0, 1, 1, 0, 1},
{1, 1, 0, 1, 0},
{0, 1, 1, 1, 0},
{1, 1, 1, 1, 0},
{1, 1, 1, 1, 1},
{0, 0, 0, 0, 0}};
printMaxSubSquare(M);
}
}PHP
$max_i - $max_of_s; $i--)
{
for($j = $max_j;
$j > $max_j - $max_of_s; $j--)
{
echo $M[$i][$j], " " ;
}
echo "\n" ;
}
}
# Driver code
$M = array(array(0, 1, 1, 0, 1),
array(1, 1, 0, 1, 0),
array(0, 1, 1, 1, 0),
array(1, 1, 1, 1, 0),
array(1, 1, 1, 1, 1),
array(0, 0, 0, 0, 0));
$R = 6 ;
$C = 5 ;
printMaxSubSquare($M, $R, $C);
// This code is contributed by Ryuga
?>输出：
Maximum size sub-matrix is:
1 1 1
1 1 1
1 1 1
时间复杂度： O(m * n)其中m是给定矩阵中的行数，n是列数。
辅助空间： O(m * n)其中m是给定矩阵中的行数，n是列数。
算法范例：动态编程