在 Q 查询的给定范围内翻转子矩阵后的二进制矩阵

给定一个维度为M x N的二进制矩阵arr[][]和形式为 ( x1, y1, x2, y2) 的Q查询，其中 ( x1, y1 ) 和 ( x2, y2 ) 表示左上和下-分别需要翻转子矩阵的右索引（将 0 转换为 1，反之亦然）。任务是打印执行给定Q查询后获得的最终矩阵。

例子：

Input: arr[][] = {{0, 1, 0}, {1, 1, 0}}, queries[][] = {{1, 1, 2, 3}}
Output: [[1, 0, 1], [0, 0, 1]]
Explanation:
The submatrix to be flipped is equal to {{0, 1, 0}, {1, 1, 0}} 
The flipped matrix is {{1, 0, 1}, {0, 0, 1}}.

Input: arr[][] = {{0, 1, 0}, {1, 1, 0}}, queries[][] = {{1, 1, 2, 3}, {1, 1, 1, 1], {1, 2, 2, 3}}
Output: [[0, 1, 0], [0, 1, 0]]
Explanation:
Query 1:
Submatrix to be flipped = [[0, 1, 0], [1, 1, 0]] 
Flipped submatrix is [[1, 0, 1], [0, 0, 1]]. 
Therefore, the modified matrix is [[1, 0, 1], [0, 0, 1]].
Query 2: 
Submatrix to be flipped = [[1]] 
Flipped submatrix is [[0]] 
Therefore, the matrix is [[0, 0, 1], [0, 0, 1]].
Query 3:
Submatrix to be flipped = [[0, 1], [0, 1]] 
Flipped submatrix is [[1, 0], [1, 0]] 
Therefore, the modified matrix is [[0, 1, 0], [0, 1, 0]].

朴素的方法：解决每个查询问题的最简单方法是迭代给定的子矩阵和每个元素，检查它是 0 还是 1，并相应地翻转。在对所有查询完成这些操作后，打印得到的最终矩阵。

下面是上述方法的实现：

C++

// C++ program to implement
// the above approach
#include
using namespace std;
 
// Function to flip a submatrices
void manipulation(vector> &matrix,
                         vector &q)
{
   
  // Boundaries of the submatrix
  int x1 = q[0], y1 = q[1],
      x2 = q[2], y2 = q[3];
 
  // Iterate over the submatrix
  for(int i = x1 - 1; i < x2; i++)
  {
    for(int j = y1 - 1; j < y2; j++)
    {
       
      // Check for 1 or 0
      // and flip accordingly
      if (matrix[i][j] == 1)
        matrix[i][j] = 0;
      else
        matrix[i][j] = 1;
    }
  }
}
 
// Function to perform the queries
void queries_fxn(vector> &matrix,
                 vector> &queries)
{
  for(auto q : queries)
    manipulation(matrix, q);       
}
 
// Driver code
int main()
{
  vector> matrix = { { 0, 1, 0 },
                                 { 1, 1, 0 } };
  vector> queries = { { 1, 1, 2, 3 },
                                  { 1, 1, 1, 1 },
                                  { 1, 2, 2, 3 } };
   
  // Function call
  queries_fxn(matrix, queries);
  cout << "[";
 
  for(int i = 0; i < matrix.size(); i++)
  {
    cout << "["; 
    for(int j = 0; j < matrix[i].size(); j++)
      cout << matrix[i][j] << " ";
 
    if (i == matrix.size() - 1)
      cout << "]";
    else
      cout << "], ";
  }
   cout << "]";
}
 
// This code is contributed by bgangwar59

Java

// Java program to implement
// the above approach
import java.util.*;
import java.lang.*;
class GFG{
 
// Function to flip a submatrices
static void manipulation(int[][] matrix,
                         int[] q)
{
  // Boundaries of the submatrix
  int x1 = q[0], y1 = q[1],
      x2 = q[2], y2 = q[3];
 
  // Iterate over the submatrix
  for(int i = x1 - 1; i < x2; i++)
  {
    for(int j = y1 - 1; j < y2; j++)
    {
      // Check for 1 or 0
      // and flip accordingly
      if (matrix[i][j] == 1)
        matrix[i][j] = 0;
      else
        matrix[i][j] = 1;
    }
  }
}
 
// Function to perform the queries
static void queries_fxn(int[][] matrix,
                        int[][] queries)
{
  for(int[] q : queries)
    manipulation(matrix, q);       
}
 
// Driver code
public static void main (String[] args)
{
  int[][] matrix = {{0, 1, 0}, {1, 1, 0}};
  int[][] queries = {{1, 1, 2, 3},
                     {1, 1, 1, 1},
                     {1, 2, 2, 3}};
 
  // Function call
  queries_fxn(matrix, queries);
  System.out.print("[");
 
  for(int i = 0; i < matrix.length; i++)
  {
    System.out.print("["); 
    for(int j = 0; j < matrix[i].length; j++)
      System.out.print(matrix[i][j] + " ");
 
    if(i == matrix.length - 1)
      System.out.print("]");
    else
      System.out.print("], ");
  }
  System.out.print("]");
}
}
 
// This code is contributed by offbeat

Python3

# Python3 Program to implement
# the above approach
 
# Function to flip a submatrices
def manipulation(matrix, q):
 
    # Boundaries of the submatrix
    x1, y1, x2, y2 = q
 
    # Iterate over the submatrix
    for i in range(x1-1, x2):
        for j in range(y1-1, y2):
 
            # Check for 1 or 0
            # and flip accordingly
            if matrix[i][j]:
                matrix[i][j] = 0
            else:
                matrix[i][j] = 1
 
# Function to perform the queries
def queries_fxn(matrix, queries):
    for q in queries:
        manipulation(matrix, q)
 
 
# Driver Code
matrix = [[0, 1, 0], [1, 1, 0]]
queries = [[1, 1, 2, 3], \
           [1, 1, 1, 1], \
           [1, 2, 2, 3]]
 
# Function call
queries_fxn(matrix, queries)
print(matrix)

C#

// C# program to implement
// the above approach
using System;
 
class GFG{
 
// Function to flip a submatrices
static void manipulation(int[,] matrix,
                         int[] q)
{
     
    // Boundaries of the submatrix
    int x1 = q[0], y1 = q[1],
        x2 = q[2], y2 = q[3];
     
    // Iterate over the submatrix
    for(int i = x1 - 1; i < x2; i++)
    {
        for(int j = y1 - 1; j < y2; j++)
        {
             
            // Check for 1 or 0
            // and flip accordingly
            if (matrix[i, j] == 1)
                matrix[i, j] = 0;
            else
                matrix[i, j] = 1;
        }
    }
}
 
public static int[] GetRow(int[,] matrix, int row)
{
    var rowLength = matrix.GetLength(1);
    var rowVector = new int[rowLength];
     
    for(var i = 0; i < rowLength; i++)
        rowVector[i] = matrix[row, i];
     
    return rowVector;
}
 
// Function to perform the queries
static void queries_fxn(int[,] matrix,
                        int[,] queries)
{
    for(int i = 0; i < queries.GetLength(0); i++)
        manipulation(matrix, GetRow(queries, i));       
}
 
// Driver code
public static void Main(String[] args)
{
    int[,] matrix = { { 0, 1, 0 },
                      { 1, 1, 0 } };
    int[,] queries = { { 1, 1, 2, 3 },
                       { 1, 1, 1, 1 },
                       { 1, 2, 2, 3 } };
     
    // Function call
    queries_fxn(matrix, queries);
    Console.Write("[");
     
    for(int i = 0; i < matrix.GetLength(0); i++)
    {
        Console.Write("["); 
        for(int j = 0; j < matrix.GetLength(1); j++)
            Console.Write(matrix[i, j] + ", ");
         
        if (i == matrix.Length - 1)
            Console.Write("]");
        else
            Console.Write("], ");
    }
    Console.Write("]");
}
}
 
// This code is contributed by Princi Singh

Javascript

// Javascript program to implement
// the above approach
 
// Function to flip a submatrices
function manipulation(matrix, q) {
 
    // Boundaries of the submatrix
    let x1 = q[0], y1 = q[1],
        x2 = q[2], y2 = q[3];
 
    // Iterate over the submatrix
    for (let i = x1 - 1; i < x2; i++) {
        for (let j = y1 - 1; j < y2; j++) {
 
            // Check for 1 or 0
            // and flip accordingly
            if (matrix[i][j] == 1)
                matrix[i][j] = 0;
            else
                matrix[i][j] = 1;
        }
    }
}
 
// Function to perform the queries
function queries_fxn(matrix, queries) {
    for (let q of queries)
        manipulation(matrix, q);
}
 
// Driver code
 
let matrix = [[0, 1, 0],
              [1, 1, 0]];
let queries = [[1, 1, 2, 3],
               [1, 1, 1, 1],
               [1, 2, 2, 3]];
 
// Function call
queries_fxn(matrix, queries);
document.write("[");
 
for (let i = 0; i < matrix.length; i++) {
    document.write("[");
    for (let j = 0; j < matrix[i].length; j++) {
        if (j < matrix[i].length - 1) {
            document.write(matrix[i][j] + ", ");
        }
        else {
            document.write(matrix[i][j] + " ");
        }
    }
 
    if (i == matrix.length - 1)
        document.write("]");
    else
        document.write("], ");
}
document.write("]");
 
 
 
// This code is contributed by _saurabh_jaiswal

输出：

[[0, 1, 0], [0, 1, 0]]

时间复杂度： O(N * M * Q)
辅助空间： O(1)

高效方法：上述方法可以使用动态规划和前缀和技术进行优化。标记每个查询涉及的子矩阵的边界，然后计算矩阵涉及的操作的前缀和，并相应地更新矩阵。请按照以下步骤解决问题：

• 初始化一个二维状态空间表dp[][]以存储矩阵的各个索引处的翻转计数
• 对于每个查询 {x1, y1, x2, y2, K}，通过以下操作更新 dp[][] 矩阵：
  • dp[x1][y1] += 1
  • dp[x2 + 1][y1] -= 1
  • dp[x2 + 1][y2 + 1] += 1
  • dp[x1][y2 + 1] -= 1
• 现在，遍历 dp[][] 矩阵并通过以下关系计算行列和对角线的前缀和来更新 dp[i][j]：

dp[i][j] = dp[i][j] + dp[i-1][j] + dp[i][j – 1] – dp[i – 1][j – 1] 
 

• 如果发现 dp[i][j] 为奇数，则将 mat[i – 1][j – 1] 减 1。
• 最后，打印更新后的矩阵 mat[][] 作为结果。

下面是上述方法的实现：

蟒蛇3

# Python3 program to implement
# the above approach
 
# Function to modify dp[][] array by
# generating prefix sum
def modifyDP(matrix, dp):
     
    for j in range(1, len(matrix)+1):
         
        for k in range(1, len(matrix[0])+1):
             
            # Update the tabular data
            dp[j][k] = dp[j][k] + dp[j-1][k] \
            + dp[j][k-1]-dp[j-1][k-1]
             
            # If the count of flips is even
            if dp[j][k] % 2 != 0:
                matrix[j-1][k-1] = int(matrix[j-1][k-1]) ^ 1
 
# Function to update dp[][] matrix
# for each query
def queries_fxn(matrix, queries, dp):
    for q in queries:
        x1, y1, x2, y2 = q
         
        # Update the table
        dp[x1][y1] += 1
        dp[x2 + 1][y2 + 1] += 1
        dp[x1][y2 + 1] -= 1
        dp[x2 + 1][y1] -= 1
         
    modifyDP(matrix, dp)
 
 
# Driver Code
matrix = [[0, 1, 0], [1, 1, 0]]
queries = [[1, 1, 2, 3], \
           [1, 1, 1, 1], \
           [1, 2, 2, 3]]
 
# Initialize dp table
dp = [[0 for i in range(len(matrix[0])+2)] \
for j in range(len(matrix)+2)]
 
queries_fxn(matrix, queries, dp)
print(matrix)

输出：

[[0, 1, 0], [0, 1, 0]]

时间复杂度： O(N * M)
辅助空间： O(N * M)

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