给定一个矩阵的N行M列，任务是使用前K个整数填充矩阵像元，使得：
- 每组单元由单个整数表示。
- 包含最大和最小单元数的组之间的差异应最小。
- 同一组的所有像元都应该是连续的，即，对于任何一组，两个相邻的像元应遵循| x i + 1 – x i |的规则。 + | y i + 1 – y i | = 1 。
例子：
Input: N = 5, M = 5, K = 6
Output:
1 1 1 1 1
3 2 2 2 2
3 3 3 4 4
5 5 5 4 4
5 6 6 6 6
Explanation:
The above matrix follows all the conditions above and dividing the matrix into K different groups.
Input: N = 2, M = 3, K = 3
Output:
1 1 2
3 3 2
Explanation:
For making three group of the matrix each should have the group of size two.
So, to reduce the difference between the group containing maximum and minimum no of cells and all the matrix cells are used to make the K different groups having all the adjacent elements of the same group follow the |xi + 1 – xi| + |yi + 1 – yi| = 1 as well.
方法：以下是步骤：
- 创建大小为N * M的矩阵。
- 为了减少包含最大和最小单元数的组之间的差异，请用至少(N * M)/ K个单元填充所有部分。
- 其余部分将包含(N * M)/ K +1个单元格。
- 要遵循给定的规则，遍历矩阵并相应地用不同的部分填充矩阵。
- 完成上述步骤后，打印矩阵。
下面是上述方法的实现：
C++
// C++ program for the above approach
#include
using namespace std;
// Function to fill the matrix with
// the given conditions
void fillMatrix(int** mat, int& row,
int& col, int sizeOfpart,
int noOfPart, int& start,
int m, int& flag)
{
// Count of parts with size sizeOfPart
for (int i = 0; i < noOfPart; ++i) {
int count = 0;
while (count < sizeOfpart) {
// Assigning the cell
// with no of groups
mat[row][col] = start;
// Update row
if (col == m - 1
&& flag == 1) {
row++;
col = m;
flag = 0;
}
else if (col == 0
&& flag == 0) {
row++;
col = -1;
flag = 1;
}
// Update col
if (flag == 1) {
col++;
}
else {
col--;
}
// Increment count
count++;
}
// For new group increment start
start++;
}
}
// Function to return the reference of
// the matrix to be filled
int** findMatrix(int N, int M, int k)
{
// Create matrix of size N*M
int** mat = (int**)malloc(
N * sizeof(int*));
for (int i = 0; i < N; ++i) {
mat[i] = (int*)malloc(
M * sizeof(int));
}
// Starting index of the matrix
int row = 0, col = 0;
// Size of one group
int size = (N * M) / k;
int rem = (N * M) % k;
// Element to assigned to matrix
int start = 1, flag = 1;
// Fill the matrix that have rem
// no of parts with size size + 1
fillMatrix(mat, row, col, size + 1,
rem, start, M, flag);
// Fill the remaining number of parts
// with each part size is 'size'
fillMatrix(mat, row, col, size,
k - rem, start, M, flag);
// Return the matrix
return mat;
}
// Function to print the matrix
void printMatrix(int** mat, int N,
int M)
{
// Traverse the rows
for (int i = 0; i < N; ++i) {
// Traverse the columns
for (int j = 0; j < M; ++j) {
cout << mat[i][j] << " ";
}
cout << endl;
}
}
// Driver Code
int main()
{
// Given N, M, K
int N = 5, M = 5, K = 6;
// Function Call
int** mat = findMatrix(N, M, K);
// Function Call to print matrix
printMatrix(mat, N, M);
return 0;
} Java
// Java program for the above approach
import java.util.*;
class GFG{
static int N, M;
static int [][]mat;
static int row, col, start, flag;
// Function to fill the matrix with
// the given conditions
static void fillMatrix(int sizeOfpart,
int noOfPart,
int m)
{
// Count of parts with size sizeOfPart
for(int i = 0; i < noOfPart; ++i)
{
int count = 0;
while (count < sizeOfpart)
{
// Assigning the cell
// with no of groups
mat[row][col] = start;
// Update row
if (col == m - 1 && flag == 1)
{
row++;
col = m;
flag = 0;
}
else if (col == 0 && flag == 0)
{
row++;
col = -1;
flag = 1;
}
// Update col
if (flag == 1)
{
col++;
}
else
{
col--;
}
// Increment count
count++;
}
// For new group increment start
start++;
}
}
// Function to return the reference of
// the matrix to be filled
static void findMatrix(int k)
{
// Create matrix of size N*M
mat = new int[M][N];
// Starting index of the matrix
row = 0;
col = 0;
// Size of one group
int size = (N * M) / k;
int rem = (N * M) % k;
// Element to assigned to matrix
start = 1;
flag = 1;
// Fill the matrix that have rem
// no of parts with size size + 1
fillMatrix(size + 1, rem, M);
// Fill the remaining number of parts
// with each part size is 'size'
fillMatrix(size, k - rem, M);
}
// Function to print the matrix
static void printMatrix()
{
// Traverse the rows
for(int i = 0; i < N; ++i)
{
// Traverse the columns
for(int j = 0; j < M; ++j)
{
System.out.print(mat[i][j] + " ");
}
System.out.println();
}
}
// Driver Code
public static void main(String[] args)
{
// Given N, M, K
N = 5;
M = 5;
int K = 6;
// Function Call
findMatrix(K);
// Function Call to print matrix
printMatrix();
}
}
// This code is contributed by Amit KatiyarC#
// C# program for the
// above approach
using System;
class GFG{
static int N, M;
static int [,]mat;
static int row, col,
start, flag;
// Function to fill the
// matrix with the given
// conditions
static void fillMatrix(int sizeOfpart,
int noOfPart,
int m)
{
// Count of parts with size
// sizeOfPart
for(int i = 0;
i < noOfPart; ++i)
{
int count = 0;
while (count < sizeOfpart)
{
// Assigning the cell
// with no of groups
mat[row, col] = start;
// Update row
if (col == m - 1 &&
flag == 1)
{
row++;
col = m;
flag = 0;
}
else if (col == 0 &&
flag == 0)
{
row++;
col = -1;
flag = 1;
}
// Update col
if (flag == 1)
{
col++;
}
else
{
col--;
}
// Increment count
count++;
}
// For new group increment
// start
start++;
}
}
// Function to return the
// reference of the matrix
// to be filled
static void findMatrix(int k)
{
// Create matrix of
// size N*M
mat = new int[M, N];
// Starting index of the
// matrix
row = 0;
col = 0;
// Size of one group
int size = (N * M) / k;
int rem = (N * M) % k;
// Element to assigned to
// matrix
start = 1;
flag = 1;
// Fill the matrix that have
// rem no of parts with size
// size + 1
fillMatrix(size + 1,
rem, M);
// Fill the remaining number
// of parts with each part
// size is 'size'
fillMatrix(size, k - rem, M);
}
// Function to print the
// matrix
static void printMatrix()
{
// Traverse the rows
for(int i = 0; i < N; ++i)
{
// Traverse the columns
for(int j = 0; j < M; ++j)
{
Console.Write(mat[i, j] +
" ");
}
Console.WriteLine();
}
}
// Driver Code
public static void Main(String[] args)
{
// Given N, M, K
N = 5;
M = 5;
int K = 6;
// Function Call
findMatrix(K);
// Function Call to
// print matrix
printMatrix();
}
}
// This code is contributed by 29AjayKumar输出：
1 1 1 1 1
3 2 2 2 2
3 3 3 4 4
5 5 5 4 4
5 6 6 6 6
时间复杂度： O(N * M)
辅助空间： O(N * M)