Submission #58139

#TimeUsernameProblemLanguageResultExecution timeMemory
58139BenqRiddick's Cube (IZhO13_riddicks)C++14
100 / 100
163 ms920 KiB

#include <bits/stdc++.h>
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/pb_ds/assoc_container.hpp>

using namespace std;
using namespace __gnu_pbds;
 
typedef long long ll;
typedef long double ld;
typedef complex<ld> cd;

typedef pair<int, int> pi;
typedef pair<ll,ll> pl;
typedef pair<ld,ld> pd;

typedef vector<int> vi;
typedef vector<ld> vd;
typedef vector<ll> vl;
typedef vector<pi> vpi;
typedef vector<pl> vpl;
typedef vector<cd> vcd;

template <class T> using Tree = tree<T, null_type, less<T>, rb_tree_tag,tree_order_statistics_node_update>;

#define FOR(i, a, b) for (int i=a; i<(b); i++)
#define F0R(i, a) for (int i=0; i<(a); i++)
#define FORd(i,a,b) for (int i = (b)-1; i >= a; i--)
#define F0Rd(i,a) for (int i = (a)-1; i >= 0; i--)

#define sz(x) (int)(x).size()
#define mp make_pair
#define pb push_back
#define f first
#define s second
#define lb lower_bound
#define ub upper_bound
#define all(x) x.begin(), x.end()

const int MOD = 1000000007;
const ll INF = 1e18;
const int MX = 100001;

int N,M,g[5][5], ret = 100500;
vi cur;

void rotCol(int ind) {
    char c = g[0][ind];
    F0R(i,N-1) g[i][ind] = g[i+1][ind];
    g[N-1][ind] = c;
}

void rotRow(int ind) {
    char c = g[ind][0];
    F0R(i,M-1) g[ind][i] = g[ind][i+1];
    g[ind][M-1] = c;
}

bool checkRows() {
    F0R(i,N) FOR(j,1,M) if (g[i][j] != g[i][0]) return 0;
    return 1;
}

bool checkCols() {
    F0R(j,M) FOR(i,1,N) if (g[i][j] != g[0][j]) return 0;
    return 1;
}

void SEARCH(int ind, int ans) {
    if (ind == N) {
        if (checkCols() || checkRows()) ret = min(ret,ans);
        return;
    }
    F0R(i,M) {
        SEARCH(ind+1,ans+min(i,M-i));
        rotRow(ind);
    }
}

void search(int ind = 0, int ans = 0) {
    if (ind == M) {
        SEARCH(0,ans);
        return;
    }
    F0R(i,N) {
        search(ind+1,ans+min(i,N-i));
        rotCol(ind);
    }
}

int main() {
    ios_base::sync_with_stdio(0); cin.tie(0);
    cin >> N >> M; cur.resize(M);
    F0R(i,N) F0R(j,M) cin >> g[i][j];
    search();
    cout << ret;
}

/* Look for:
* the exact constraints (multiple sets are too slow for n=10^6 :( ) 
* special cases (n=1?)
* overflow (ll vs int?)
* array bounds
*/
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