给定两个大小为M×N的矩阵A [] []和B [] []和整数X ，任务是检查是否可以通过加法将矩阵A [] []转换为矩阵B [] []在同一行或同一列中的X个连续单元的任何值任意次数(可能为零)。
例子：
Input: A[][] = { {0, 0}, {0, 0}}, B[][] = {{1, 2}, {0, 1}}, X = 2
Output: Yes
Explanation:
Operation 1: Addinng 1 to A[0][0] and A[0][1] modifies A[][] to {{1, 1}, {0, 0}}.
Operation 2: Adding 1 to A[0][1] and A[1][1] modifies A[][] to {{1, 2}, {0, 1}}.
After performing this two operations, matrix A[][] and B[][] are equal.
Input: A= {{0, 0, 0}, {0, 0, 0}}, B = {{1, 2, 3}, {4, 5, 6}}, X = 4
Output: False
方法：可以通过先执行所有水平操作再执行垂直操作来贪婪地解决问题。
请按照以下步骤解决问题：
- 使用变量i和j在[0，M – 1]和[0，N – X]范围内遍历矩阵以执行水平运算，并执行以下运算：
- 如果A [i] [j]不等于B [i] [j] ，则将同一行中的下X个元素增加A [i] [j] – B [i] [j] 。
- 现在，遍历矩阵以使用[0，M – X]和[0，N – 1]范围内的变量i和j进行垂直运算，并执行以下运算：
- 检查A [i] [j]是否等于B [i] [j] 。
- 如果发现为假，则将同一列中的下一个X元素增加A [i] [j] – B [i] [j] 。
- 如果矩阵A [] []和B [] []相等，则打印“ True” 。否则，打印“ False” 。
下面是上述方法的实现：
C++
// C++ Program for the above approach
#include
using namespace std;
// Function to check whether Matrix A[][]
// can be transformed to Matrix B[][] or not
bool Check(int A[][2], int B[][2],
int M, int N, int X)
{
// Traverse the matrix to perform
// horizontal operations
for (int i = 0; i < M; i++) {
for (int j = 0; j <= N - X; j++) {
if (A[i][j] != B[i][j]) {
// Calculate difference
int diff = B[i][j] - A[i][j];
for (int k = 0; k < X; k++) {
// Update next X elements
A[i][j + k] = A[i][j + k] + diff;
}
}
}
}
// Traverse the matrix to perform
// vertical operations
for (int i = 0; i <= M - X; i++) {
for (int j = 0; j < N; j++) {
if (A[i][j] != B[i][j]) {
// Calculate difference
int diff = B[i][j] - A[i][j];
for (int k = 0; k < X; k++) {
// Update next K elements
A[i + k][j] = A[i + k][j] + diff;
}
}
}
}
for (int i = 0; i < M; i++) {
for (int j = 0; j < N; j++) {
// A[i][j] is not equal to B[i][j]
if (A[i][j] != B[i][j]) {
// Conversion is not possible
return 0;
}
}
}
// Conversion is possible
return 1;
}
// Driver Code
int main()
{
// Input
int M = 2, N = 2, X = 2;
int A[2][2] = { { 0, 0 }, { 0, 0 } };
int B[2][2] = { { 1, 2 }, { 0, 1 } };
if (Check(A, B, M, N, X)) {
cout << "Yes" << endl;
}
else {
cout << "No" << endl;
}
return 0;
} Java
// Java program for the above approach
import java.io.*;
import java.util.*;
class GFG{
// Function to check whether Matrix A[][]
// can be transformed to Matrix B[][] or not
static int Check(int A[][], int B[][],
int M, int N, int X)
{
// Traverse the matrix to perform
// horizontal operations
for(int i = 0; i < M; i++)
{
for(int j = 0; j <= N - X; j++)
{
if (A[i][j] != B[i][j])
{
// Calculate difference
int diff = B[i][j] - A[i][j];
for(int k = 0; k < X; k++)
{
// Update next X elements
A[i][j + k] = A[i][j + k] + diff;
}
}
}
}
// Traverse the matrix to perform
// vertical operations
for(int i = 0; i <= M - X; i++)
{
for(int j = 0; j < N; j++)
{
if (A[i][j] != B[i][j])
{
// Calculate difference
int diff = B[i][j] - A[i][j];
for(int k = 0; k < X; k++)
{
// Update next K elements
A[i + k][j] = A[i + k][j] + diff;
}
}
}
}
for(int i = 0; i < M; i++)
{
for(int j = 0; j < N; j++)
{
// A[i][j] is not equal to B[i][j]
if (A[i][j] != B[i][j])
{
// Conversion is not possible
return 0;
}
}
}
// Conversion is possible
return 1;
}
// Driver Code
public static void main(String[] args)
{
// Input
int M = 2, N = 2, X = 2;
int A[][] = { { 0, 0 }, { 0, 0 } };
int B[][] = { { 1, 2 }, { 0, 1 } };
if (Check(A, B, M, N, X) != 0)
{
System.out.println("Yes");
}
else
{
System.out.println("No");
}
}
}
// This code is contributed by sanjoy_62Python3
# Python3 program for the above approach
# Function to check whether Matrix A[][]
# can be transformed to Matrix B[][] or not
def Check(A, B, M, N, X):
# Traverse the matrix to perform
# horizontal operations
for i in range(M):
for j in range(N - X + 1):
if (A[i][j] != B[i][j]):
# Calculate difference
diff = B[i][j] - A[i][j]
for k in range(X):
# Update next X elements
A[i][j + k] = A[i][j + k] + diff
# Traverse the matrix to perform
# vertical operations
for i in range(M - X + 1):
for j in range(N):
if (A[i][j] != B[i][j]):
# Calculate difference
diff = B[i][j] - A[i][j]
for k in range(X):
# Update next K elements
A[i + k][j] = A[i + k][j] + diff
for i in range(M):
for j in range(N):
# A[i][j] is not equal to B[i][j]
if (A[i][j] != B[i][j]):
# Conversion is not possible
return 0
# Conversion is possible
return 1
# Driver Code
if __name__ == '__main__':
# Input
M, N, X = 2, 2, 2
A = [ [ 0, 0 ], [ 0, 0 ] ]
B = [ [ 1, 2 ], [ 0, 1 ] ]
if (Check(A, B, M, N, X)):
print("Yes")
else:
print("No")
# This code is contributed by mohit kumar 29C#
// C# program for the above approach
using System;
using System.Collections.Generic;
class GFG{
// Function to check whether Matrix A[][]
// can be transformed to Matrix B[][] or not
static int Check(int [,]A, int [,]B,
int M, int N, int X)
{
// Traverse the matrix to perform
// horizontal operations
for(int i = 0; i < M; i++)
{
for(int j = 0; j <= N - X; j++)
{
if (A[i, j] != B[i, j])
{
// Calculate difference
int diff = B[i, j] - A[i, j];
for(int k = 0; k < X; k++)
{
// Update next X elements
A[i, j + k] = A[i, j + k] + diff;
}
}
}
}
// Traverse the matrix to perform
// vertical operations
for(int i = 0; i <= M - X; i++)
{
for(int j = 0; j < N; j++)
{
if (A[i, j] != B[i, j])
{
// Calculate difference
int diff = B[i,j] - A[i,j];
for(int k = 0; k < X; k++)
{
// Update next K elements
A[i + k, j] = A[i + k, j] + diff;
}
}
}
}
for(int i = 0; i < M; i++)
{
for(int j = 0; j < N; j++)
{
// A[i][j] is not equal to B[i][j]
if (A[i, j] != B[i, j])
{
// Conversion is not possible
return 0;
}
}
}
// Conversion is possible
return 1;
}
// Driver Code
public static void Main()
{
// Input
int M = 2, N = 2, X = 2;
int [,]A = { { 0, 0 }, { 0, 0 } };
int [,]B = { { 1, 2 }, { 0, 1 } };
if (Check(A, B, M, N, X) == 1)
{
Console.WriteLine("Yes");
}
else
{
Console.WriteLine("No");
}
}
}
// This code is contributed by SURENDRA_GANGWAR输出：
Yes时间复杂度： O(M * N * X)
辅助空间： O(1)