1294E - Obtain a Permutation - CodeForces Solution

#include<bits/stdc++.h>

#define ll long long
#define pp push_back
#define endl '\n'
#define all(x) x.begin(),x.end()
#define ld long double
#define PI acos(-1)
#define sin(a) sin((a)*PI/180)
#define cos(a) cos((a)*PI/180)
#define ones(x) __builtin_popcountll(x)
//#define int ll

#define Drakon  ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);
using namespace std;


unsigned long long inf = 1e10;
const double EPS = 1e-6;
const int MOD = 1000000007, N = 200005, LOG = 25, M = 2;

typedef array<array<int, M>, M> matrix;

ll gcd(ll x, ll y) {
    return y ? gcd(y, x % y) : x;
}

ll lcm(ll a, ll b) {
    return (a * b) / __gcd(a, b);
}

void solve() {
    int n,m;
    cin >> n >> m;
    int arr[n][m],need[n][m];
    for (int i = 0; i < n; ++i) {
        for (int j = 0; j < m; ++j) {
            cin >> arr[i][j];
            need[i][j] = i * m + j + 1;
        }
    }
    int ans = 0;
    for (int j = 0; j < m; ++j) {
        vector<int>cur,want;
        for (int i = 0; i < n; ++i) {
            cur.pp(arr[i][j]);
            want.pp(need[i][j]);
        }
        map<int,int>pos;
        for(int i = 0; i < n; i ++){
            pos[want[i]] = i;
        }
        int cycle[n];
        memset(cycle,0,sizeof cycle);
        for (int i = 0; i < n; ++i) {
            if(pos.count(cur[i])){
                int tmp = (i - pos[cur[i]] + n) % n;
                cycle[tmp] ++;
            }
        }
        int ret = 1e9;
        for (int i = 0; i < n; ++i) {
            ret = min(ret,n - cycle[i] + i);
        }
        ans += ret;
    }
    cout << ans << endl;
}

signed main() {
    Drakon
    int t = 1;
//    cin >> t;
    while (t--) {
        solve();
    }
}