AcWing 1874. Moo encryption (enumeration, hash)

【 Title Description 】
Many people don't know , Cows love jigsaw puzzles , Especially word puzzles .
Farmer John's cow recently invented an interesting “ Word Finder ” Puzzle .
An example of this puzzle is ：

USOPEN
OOMABO
MOOMXO
PQMROM

As a cow , The only words they are interested in are MOO, It can move horizontally several times in the puzzle 、 vertical 、$45$ Degree slash or $135$ Degree slash appears .
In the above example ,MOO All in all $6$ Time .
Farmer John is also a puzzle fan , Because the cows don't want him to solve it before them “ Word Finder ” Puzzle , So they use ” Replace password “ The content is encrypted .
The “ Replace password ” Replaced each letter in the alphabet with a different letter .
for example ,$A$ Can be mapped to $X$,$B$ Can be mapped to $A$, And so on .
No letters map to themselves , Nor do two letters map to the same letter （ Otherwise, the solution will be ambiguous ）.
Unfortunately , The cow forgot the specific encryption method of replacing the password .
Please help them determine if decryption with the appropriate replacement password , The biggest possible problem in the puzzle MOO Count .

【 Input format 】
The first line contains $N,M$, Indicates that the size of the puzzle is $N$ That's ok $M$ Column .
Next $N$ That's ok , Each row contains $M$ Capital letters , Describe the encrypted puzzle .

【 Output format 】
Output if decrypted with the appropriate replacement password , The biggest possible problem in the puzzle MOO Count .

【 Data range 】
$1\le N,M\le 50$

【 sample input 】

4 6
TAMHGI
MMQVWM
QMMQSM
HBQUMQ

【 sample output 】

6

【 Sample explanation 】
In this example ,$M$ and $O$ Are replaced by $Q$ and $M$.
Therefore, decryption can exist at most $6$ individual MOO.

【 analysis 】

We can map any letter to another letter , But it can't be mapped to itself , So all ABB Type and not MOO All strings can be mapped to MOO.

Enumeration starts at each point , Then enumerate the length that can be constructed by extending in eight directions $3$ String , If the constructed string $s$ Form like ABB, A string that is not the first letter of M, The last two letters are not O, Then put the string $s$ The number of occurrences plus one . Finally, the most frequent occurrence of all qualified strings is the answer .

【 Code 】

#include <iostream>
#include <cstring>
#include <algorithm>
#include <string>
#include <unordered_map>
using namespace std;

const int N = 60;
char g[N][N];
unordered_map<string, int> cnt;// All in the shape of ABB And A Not for M、B Not for O The number of strings 
int n, m;
int dx[8] = {
     -1, -1, -1, 0, 1, 1, 1, 0 };
int dy[8] = {
     -1, 0, 1, 1, 1, 0, -1, -1 };

int main()
{
    
    cin >> n >> m;
    for (int i = 0; i < n; i++) cin >> g[i];
    for (int i = 0; i < n; i++)
        for (int j = 0; j < m; j++)
        {
    
            for (int d = 0; d < 8; d++)
            {
    
                string s;
                s += g[i][j];
                int x = i, y = j;
                bool flag = true;// If you cross the line, then flag by false
                for (int u = 0; u < 2; u++)
                {
    
                    x += dx[d], y += dy[d];
                    if (x < 0 || x >= n || y < 0 || y >= m) {
     flag = false; break; }
                    s += g[x][y];
                }
                if (flag && s[0] != s[1] && s[1] == s[2] && s[0] != 'M' && s[1] != 'O') cnt[s]++;
            }
        }
    int res = 0;
    for (auto& [k, v] : cnt) res = max(res, v);
    cout << res << endl;
    return 0;
}