给定一个仅包含0和1的N * M矩阵,任务是计算包含所有1的正方形子矩阵的数量。
例子:
Input: arr[][] = {{0, 1, 1, 1},
{1, 1, 1, 1},
{0, 1, 1, 1}}
Output: 15
Explanation:
There are 10 squares of side length 1.
There are 4 squares of side length 2.
There is 1 square of side length 3.
Total number of squares = 10 + 4 + 1 = 15.
Input: arr[][] = {{1, 0, 1},
{1, 1, 0},
{1, 1, 0}}
Output: 7
方法:可以使用动态编程解决此问题。
- 令数组arr [i] [j]存储以(i,j)结尾的平方矩阵的数量
- 找出以(i,j)结尾的平方数的递归关系可以由下式给出:
- 如果arr [i] [j]为1:
- arr [i] [j] = min(min(arr [i-1] [j],arr [i] [j-1]),arr [i-1] [j-1])+ 1
- 否则,如果arr [i] [j]为0:
- arr [i] [j] = 0
- 如果arr [i] [j]为1:
- 计算该数组的总和,该总和等于全为1的正方形子矩阵的数量。
下面是上述方法的实现:
CPP
// C++ program to return the number of
// square submatrices with all 1s
#include
using namespace std;
#define n 3
#define m 3
// Function to return the number of
// square submatrices with all 1s
int countSquareMatrices(int a[][m],
int N, int M)
{
// Initialize count variable
int count = 0;
for (int i = 1; i < N; i++) {
for (int j = 1; j < M; j++) {
// If a[i][j] is equal to 0
if (a[i][j] == 0)
continue;
// Calculate number of
// square submatrices
// ending at (i, j)
a[i][j] = min(min(a[i - 1][j],
a[i][j - 1]),
a[i - 1][j - 1])
+ 1;
}
}
// Calculate the sum of the array
for (int i = 0; i < N; i++)
for (int j = 0; j < M; j++)
count += a[i][j];
return count;
}
// Driver code
int main()
{
int arr[][m] = { { 1, 0, 1 },
{ 1, 1, 0 },
{ 1, 1, 0 } };
cout << countSquareMatrices(arr, n, m);
return 0;
}
Java
// Java program to return the number of
// square submatrices with all 1s
class GFG
{
final static int n = 3;
final static int m = 3;
// Function to return the number of
// square submatrices with all 1s
static int countSquareMatrices(int a[][], int N, int M)
{
// Initialize count variable
int count = 0;
for (int i = 1; i < N; i++)
{
for (int j = 1; j < M; j++)
{
// If a[i][j] is equal to 0
if (a[i][j] == 0)
continue;
// Calculate number of
// square submatrices
// ending at (i, j)
a[i][j] = Math.min(Math.min(a[i - 1][j], a[i][j - 1]),
a[i - 1][j - 1]) + 1;
}
}
// Calculate the sum of the array
for (int i = 0; i < N; i++)
for (int j = 0; j < M; j++)
count += a[i][j];
return count;
}
// Driver code
public static void main (String[] args)
{
int arr[][] = { { 1, 0, 1 },
{ 1, 1, 0 },
{ 1, 1, 0 } };
System.out.println(countSquareMatrices(arr, n, m));
}
}
// This code is contributed by AnkitRai01
Python
# Python3 program to return the number of
# square submatrices with all 1s
n = 3
m = 3
# Function to return the number of
# square submatrices with all 1s
def countSquareMatrices(a, N, M):
# Initialize count variable
count = 0
for i in range(1, N):
for j in range(1, M):
# If a[i][j] is equal to 0
if (a[i][j] == 0):
continue
# Calculate number of
# square submatrices
# ending at (i, j)
a[i][j] = min([a[i - 1][j],
a[i][j - 1], a[i - 1][j - 1]])+1
# Calculate the sum of the array
for i in range(N):
for j in range(M):
count += a[i][j]
return count
# Driver code
arr = [ [ 1, 0, 1],
[ 1, 1, 0 ],
[ 1, 1, 0 ] ]
print(countSquareMatrices(arr, n, m))
# This code is contributed by mohit kumar 29
C#
// C# program to return the number of
// square submatrices with all 1s
using System;
class GFG
{
static int n = 3;
static int m = 3;
// Function to return the number of
// square submatrices with all 1s
static int countSquareMatrices(int [,]a, int N, int M)
{
// Initialize count variable
int count = 0;
for (int i = 1; i < N; i++)
{
for (int j = 1; j < M; j++)
{
// If a[i][j] is equal to 0
if (a[i, j] == 0)
continue;
// Calculate number of
// square submatrices
// ending at (i, j)
a[i, j] = Math.Min(Math.Min(a[i - 1, j], a[i, j - 1]),
a[i - 1, j - 1]) + 1;
}
}
// Calculate the sum of the array
for (int i = 0; i < N; i++)
for (int j = 0; j < M; j++)
count += a[i, j];
return count;
}
// Driver code
public static void Main()
{
int [,]arr = { { 1, 0, 1 },
{ 1, 1, 0 },
{ 1, 1, 0 } };
Console.WriteLine(countSquareMatrices(arr, n, m));
}
}
// This code is contributed by AnkitRai01
输出 :
7
时间复杂度: O(N * M)
辅助空间: O(1)