给定尺寸为N * M的矩阵M [] []，任务是找到将矩阵转换为回文矩阵所需的最少1的矩阵元素增量。
A palindrome matrix is a matrix in which every row and column is a palindrome.
例子：
Input: N = 4, M = 2, arr[][]={{5, 3}, {3, 5}, {5, 3}, {3, 5}}
Output: 8
Explanation: The palindromic matrix will be arr[][] = {{5, 5}, {5, 5}, {5, 5}, {5, 5}}
Input: N = 3, M = 3, arr[][]={{1, 2, 1}, {3, 4, 1}, {1, 2, 1}}
Output: 2
Explanation:
The palindromic matrix will be arr[][] = {{1, 2, 1}, {3, 4, 3}, {1, 2, 1}}
方法：如果arr [0] [0]的值等于X，则arr [M-1] [0] ， arr [0] [M-1]和arr [N-1] [M- 1]的回文性质也必须等于X。所有元素arr [i] [j]，arr [N – i – 1] [j]，arr [N – i – 1] [M – j – 1]，arr [i] [M – j – 1] 。因此，问题减少到找到可以以最小的增量从有关的四元组获得的数量。请按照以下步骤解决问题：
- 将矩阵分成4个象限。从(0，0)索引遍历到(((N + 1)/ 2)-1，((M + 1)/ 2)-1)(仅在第一象限中)遍历矩阵。
- 对于每个索引(i，j)，存储( i，j)，(N – i – 1，j)，(N – i – 1，M – j – 1)，(i，M – j – 1)索引集合，以便仅唯一索引将出现在集合中。
- 然后将这些唯一索引中存在的元素存储在向量值中，并评估该向量中的最大值。
- 现在，将最大值与值向量其余元素之间的差相加并更新ans。
下面是上述方法的实现：
C++
// C++ program for the above approach
#include
using namespace std;
// Function to evaluate minimum number
// of operation required to convert
// the matrix to a palindrome matrix
int palindromeMatrix(int N, int M, vector > arr)
{
// Variable to store number
// of operations required
int ans = 0;
// Iterate over the first
// quadrant of the matrix
for (int i = 0; i < (N + 1) / 2; i++) {
for (int j = 0; j < (M + 1) / 2; j++) {
// Store positions of all four
// values from four quadrants
set > s;
s.insert({ i, j });
s.insert({ i, M - j - 1 });
s.insert({ N - i - 1, j });
s.insert({ N - i - 1, M - j - 1 });
// Store the values having
// unique indexes
vector values;
for (pair p : s) {
values.push_back(
arr[p.first][p.second]);
}
// Largest value in the values vector
int max = *max_element(
values.begin(),
values.end());
// Evaluate minimum increments
// required to make all vector
// elements equal
for (int k = 0; k < values.size(); k++) {
ans += max - values[k];
}
}
}
// Print the answer
cout << ans;
}
// Driver Code
int main()
{
int N = 3, M = 3;
vector > arr
= { { 1, 2, 1 },
{ 3, 4, 1 },
{ 1, 2, 1 } };
// Function Call
palindromeMatrix(N, M, arr);
return 0;
} Java
// Java program for the
// above approach
import java.util.*;
class GFG{
static class pair
{
int first, second;
public pair(int first,
int second)
{
this.first = first;
this.second = second;
}
}
// Function to evaluate minimum number
// of operation required to convert
// the matrix to a palindrome matrix
static void palindromeMatrix(int N, int M,
int[][] arr)
{
// Variable to store number
// of operations required
int ans = 0;
// Iterate over the first
// quadrant of the matrix
for (int i = 0;
i < (N + 1) / 2; i++)
{
for (int j = 0;
j < (M + 1) / 2; j++)
{
// Store positions of all four
// values from four quadrants
HashSet s =
new HashSet<>();
s.add(new pair(i, j));
s.add(new pair(i, M - j - 1));
s.add(new pair(N - i - 1, j));
s.add(new pair(N - i - 1,
M - j - 1));
// Store the values having
// unique indexes
Vector values =
new Vector<>();
for (pair p : s)
{
values.add(
arr[p.first][p.second]);
}
// Largest value in the
// values vector
int max =
Collections.max(values);
// Evaluate minimum increments
// required to make all vector
// elements equal
for (int k = 1;
k < values.size(); k++)
{
ans += max - values.get(k);
}
}
}
// Print the answer
System.out.print(ans);
}
// Driver Code
public static void main(String[] args)
{
int N = 3, M = 3;
int[][] arr = {{1, 2, 1},
{3, 4, 1},
{1, 2, 1}};
// Function Call
palindromeMatrix(N, M, arr);
}
}
// This code is contributed by Rajput-Ji Python3
# Python3 program for the above approach
# Function to evaluate minimum number
# of operation required to convert
# the matrix to a palindrome matrix
def palindromeMatrix(N, M, arr):
# Variable to store number
# of operations required
ans = 0
# Iterate over the first
# quadrant of the matrix
for i in range((N + 1) // 2):
for j in range((M + 1) // 2):
# Store positions of all four
# values from four quadrants
s = {}
s[(i, j)] = 1
s[(i, M - j - 1)] = 1
s[(N - i - 1, j)] = 1
s[(N - i - 1, M - j - 1)] = 1
# Store the values having
# unique indexes
values = []
for p, q in s:
values.append(arr[p][q])
# Largest value in the values vector
maxm = max(values)
# Evaluate minimum increments
# required to make all vector
# elements equal
for k in range(len(values)):
ans += maxm - values[k]
# Print the answer
print(ans)
# Driver Code
if __name__ == '__main__':
N, M = 3, 3
arr = [ [ 1, 2, 1 ],
[ 3, 4, 1 ],
[ 1, 2, 1 ] ]
# Function Call
palindromeMatrix(N, M, arr)
# This code is contributed by mohit kumar 29C#
// C# program for the
// above approach
using System;
using System.Collections.Generic;
class GFG{
public class pair
{
public int first, second;
public pair(int first,
int second)
{
this.first = first;
this.second = second;
}
}
// Function to evaluate minimum number
// of operation required to convert
// the matrix to a palindrome matrix
static void palindromeMatrix(int N, int M,
int[,] arr)
{
// Variable to store number
// of operations required
int ans = 0;
// Iterate over the first
// quadrant of the matrix
for(int i = 0;
i < (N + 1) / 2; i++)
{
for(int j = 0;
j < (M + 1) / 2; j++)
{
// Store positions of all four
// values from four quadrants
HashSet s = new HashSet();
s.Add(new pair(i, j));
s.Add(new pair(i, M - j - 1));
s.Add(new pair(N - i - 1, j));
s.Add(new pair(N - i - 1,
M - j - 1));
// Store the values having
// unique indexes
List values = new List();
foreach (pair p in s)
{
values.Add(arr[p.first, p.second]);
}
// Largest value in the
// values vector
values.Sort();
int max = values[values.Count - 1];
// Evaluate minimum increments
// required to make all vector
// elements equal
for(int k = 1;
k < values.Count; k++)
{
ans += max - values[k];
}
}
}
// Print the answer
Console.Write(ans);
}
// Driver Code
public static void Main(String[] args)
{
int N = 3, M = 3;
int[,] arr = { { 1, 2, 1 },
{ 3, 4, 1 },
{ 1, 2, 1 } };
// Function Call
palindromeMatrix(N, M, arr);
}
}
// This code is contributed by Amit Katiyar 输出：
2时间复杂度： O(N * M)
辅助空间： O(1)