给定两个大小为M × N 的矩阵A[][]和B[][]以及一个整数X ，任务是检查是否可以通过添加将矩阵A[][]转换为矩阵B[][]任意次数(可能为零)在同一行或同一列中的X个连续单元格的任何值。
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：这个问题可以通过先执行所有水平操作然后执行垂直操作来贪婪地解决。
请按照以下步骤解决问题：
- 遍历矩阵以执行水平操作，使用范围[0, M – 1]和[0, N – X] 上的变量i和j ，并执行以下操作：
- 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，遍历矩阵以执行垂直操作，使用范围[0, M – X]和[0, N – 1] 上的变量i和j ，并执行以下操作：
- 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X 个元素增加A[i][j] – B[i][j] 。
- 如果矩阵 A[][] 和 B[][] 相等，则打印“True” 。否则，打印“假” 。
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_GANGWARJavascript
输出：
Yes时间复杂度： O(M * N * X)
辅助空间： O(1)
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