给定R行和C列的二进制矩阵。我们可以翻转到任意大小的子矩阵。翻转意味着将1更改为0，将0更改为1。任务是使矩阵中的1s数量最大化。输出最大数量为1s。
例子：
Input : R = 3, C =3
mat[][] = { 0, 0, 1,
0, 0, 1,
1, 0, 1 }
Output : 8
Flip
0 0 1
0 0 1
1 0 1
to get
1 1 1
1 1 1
0 1 1
Input : R = 2, C = 3
mat[][] = { 0, 0, 0,
0, 0, 0 }
Output : 6
创建一个由R行和C列组成的矩阵ones [] []，该子矩阵通过以下方式预先计算子矩阵中从(0，0)到(i，j)的个数
// Common elements in ones[i-1][j] and
// ones[i][j-1] are ones[i-1][j-1]
ones[i][j] = ones[i-1][j] + ones[i][j-1] -
ones[i-1][j-1] + (mat[i][j] == 1)
从那时起，我们只允许翻转一次子矩阵。我们将每个像元(i，j)的所有可能大小的所有可能子矩阵迭代为(i + k – 1，i + k – 1)。在将数字填充到所选子矩阵中之后，我们计算位数。
将子矩阵(i，j)翻转为(i + k – 1)后，最终矩阵中的1的总数将为整个矩阵中的1 –所选子矩阵中的1 +所选子矩阵中的0。结果是：-
一个[R] [C] – cal(i，j，i + k，j + k – 1)+ k * k – cal(i，j，i + k – 1，j + k – 1)
其中cal(a，b，c，d)表示长度为c – a的正方形子矩阵中的个数。
现在可以通过以下方式定义cal(x1，y1，x2，y2)：
一个[x2] [y2] –一个[x2] [y1 – 1] –一个[x1 – 1] [y2] +一个[x1 – 1] [y1 – 1]。
以下是此方法的实现：
C++
// C++ program to find maximum number of ones after
// one flipping in Binary Matrix
#include
#define R 3
#define C 3
using namespace std;
// Return number of ones in square submatrix of size
// k x k starting from (x, y)
int cal(int ones[R + 1][C + 1], int x, int y, int k)
{
return ones[x + k - 1][y + k - 1] - ones[x - 1][y + k - 1]
- ones[x + k - 1][y - 1] + ones[x - 1][y - 1];
}
// Return maximum number of 1s after flipping a submatrix
int sol(int mat[R][C])
{
int ans = 0;
// Precomputing the number of 1s
int ones[R + 1][C + 1] = {0};
for (int i = 1; i <= R; i++)
for (int j = 1; j <= C; j++)
ones[i][j] = ones[i - 1][j] + ones[i][j - 1] -
ones[i - 1][j - 1] +
(mat[i - 1][j - 1] == 1);
// Finding the maximum number of 1s after flipping
for (int k = 1; k <= min(R, C); k++)
for (int i = 1; i + k - 1 <= R; i++)
for (int j = 1; j + k - 1 <= C; j++)
ans = max(ans, (ones[R][C] + k * k -
2 * cal(ones, i, j, k)));
return ans;
}
// Driver code
int main()
{
int mat[R][C] = {{0, 0, 1},
{ 0, 0, 1},
{ 1, 0, 1 }
};
cout << sol(mat) << endl;
return 0;
} Java
// Java program to find maximum number of ones after
// one flipping in Binary Matrix
class GFG {
static final int R = 3;
static final int C = 3 ;
// Return number of ones in square submatrix of size
// k x k starting from (x, y)
static int cal(int ones[][], int x, int y, int k)
{
return ones[x + k - 1][y + k - 1] - ones[x - 1][y + k - 1]
- ones[x + k - 1][y - 1] + ones[x - 1][y - 1];
}
// Return maximum number of 1s after flipping a submatrix
static int sol(int mat[][])
{
int ans = 0;
int val =0;
// Precomputing the number of 1s
int ones[][] = new int[R + 1][C + 1];
for (int i = 1; i <= R; i++)
for (int j = 1; j <= C; j++) {
if(mat[i - 1][j - 1] == 1)
val=1;
ones[i][j] = ones[i - 1][j] + ones[i][j - 1] -
ones[i - 1][j - 1] +
(val);
}
// Finding the maximum number of 1s after flipping
for (int k = 1; k <= Math.min(R, C); k++)
for (int i = 1; i + k - 1 <= R; i++)
for (int j = 1; j + k - 1 <= C; j++)
ans = Math.max(ans, (ones[R][C] + k * k -
2 * cal(ones, i, j, k)));
return ans;
}
// Driver code
static public void main(String[] args) {
int mat[][] = {{0, 0, 1},
{ 0, 0, 1},
{ 1, 0, 1 }
};
System.out.println(sol(mat));
}
}
// This code is contributed by Rajput-JiPython3
# Python 3 program to find maximum number of
# ones after one flipping in Binary Matrix
R = 3
C = 3
# Return number of ones in square submatrix
# of size k x k starting from (x, y)
def cal(ones, x, y, k):
return (ones[x + k - 1][y + k - 1] -
ones[x - 1][y + k - 1] -
ones[x + k - 1][y - 1] +
ones[x - 1][y - 1])
# Return maximum number of 1s after
# flipping a submatrix
def sol(mat):
ans = 0
# Precomputing the number of 1s
ones = [[0 for i in range(C + 1)]
for i in range(R + 1)]
for i in range(1, R + 1, 1):
for j in range(1, C + 1, 1):
ones[i][j] = (ones[i - 1][j] + ones[i][j - 1] -
ones[i - 1][j - 1] +
(mat[i - 1][j - 1] == 1))
# Finding the maximum number of 1s
# after flipping
for k in range(1, min(R, C) + 1, 1):
for i in range(1, R - k + 2, 1):
for j in range(1, C - k + 2, 1):
ans = max(ans, (ones[R][C] + k * k - 2 *
cal(ones, i, j, k)))
return ans
# Driver code
if __name__ == '__main__':
mat = [[0, 0, 1],
[0, 0, 1],
[1, 0, 1]]
print(sol(mat))
# This code is contributed by
# Sahil_ShelangiaC#
// C# program to find maximum number of ones after
// one flipping in Binary Matrix
using System;
public class GFG {
static readonly int R = 3;
static readonly int C = 3 ;
// Return number of ones in square submatrix of size
// k x k starting from (x, y)
static int cal(int [,]ones, int x, int y, int k)
{
return ones[x + k - 1,y + k - 1] - ones[x - 1,y + k - 1]
- ones[x + k - 1,y - 1] + ones[x - 1,y - 1];
}
// Return maximum number of 1s after flipping a submatrix
static int sol(int [,]mat)
{
int ans = 0;
int val =0;
// Precomputing the number of 1s
int [,]ones = new int[R + 1,C + 1];
for (int i = 1; i <= R; i++)
for (int j = 1; j <= C; j++) {
if(mat[i - 1,j - 1] == 1)
val=1;
ones[i,j] = ones[i - 1,j] + ones[i,j - 1] -
ones[i - 1,j - 1] +
(val);
}
// Finding the maximum number of 1s after flipping
for (int k = 1; k <= Math.Min(R, C); k++)
for (int i = 1; i + k - 1 <= R; i++)
for (int j = 1; j + k - 1 <= C; j++)
ans = Math.Max(ans, (ones[R,C] + k * k -
2 * cal(ones, i, j, k)));
return ans;
}
// Driver code
static public void Main() {
int [,]mat = {{0, 0, 1},
{ 0, 0, 1},
{ 1, 0, 1 }
};
Console.WriteLine(sol(mat));
}
}
// This code is contributed by 29AjayKumarPHP
输出：
8
时间复杂度： O(R * C * min(R，C))