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, 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, 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 offbeatPython3
# 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 SinghPython3
# 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 * 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和列以及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)
如果您想与行业专家一起参加直播课程，请参阅Geeks Classes Live