AcWing 1944. Record keeping (hash, STL)

【 Title Description 】
Farmer John has been keeping a detailed record of the cows milking in the barn .
Every hour, a group of three cows will enter the barn to produce milk , John will record their names .
for example , stay $5$ Within hours , He may record the following list , Each row corresponds to a group of cattle entering the cowshed ：

BESSIE ELSIE MATILDA
FRAN BESSIE INGRID
BESSIE ELSIE MATILDA
MATILDA INGRID FRAN
ELSIE BESSIE MATILDA

John found that the same group of cows may appear on his list many times , In the example above ,BESSIE ELSIE MATILDA This combination appeared $3$ Time （ Although John doesn't have to record their names in the same order every time ）.
Please help John calculate the number of cattle entering the cowshed the most .

【 Input format 】
The first line contains integers $N$.
Next $N$ That's ok , Each line contains the names of a group of three cows , Each name has a length of $1\sim 10$ A string of uppercase letters between .

【 Output format 】
Output the number of cattle entering the cowshed the most .

【 Data range 】
$1\le N\le 1000$

【 sample input 】

5
BESSIE ELSIE MATILDA
FRAN BESSIE INGRID
BESSIE ELSIE MATILDA
MATILDA INGRID FRAN
ELSIE BESSIE MATILDA

【 sample output 】

3

【 analysis 】

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First, sort each set of strings , Then the same three strings in different order must be the only sequence after sorting , Store this sequence in vector<string> in . And then we use map<vector<string>, int> Count the occurrence times of each group of sequences .

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【 Code 】

#include <iostream>
#include <cstring>
#include <algorithm>
#include <vector>
#include <map>
using namespace std;

map<vector<string>, int> cnt;
int n, res;

int main()
{
    
    cin >> n;
    while (n--)
    {
    
        vector<string> str(3);
        for (int i = 0; i < 3; i++) cin >> str[i];
        sort(str.begin(), str.end());
        cnt[str]++;
    }
    for (auto &[k, v] : cnt) res = max(res, v);
    cout << res << endl;
    return 0;
}