Submission #219412

#	Time	Username	Problem	Language	Result	Execution time	Memory
219412		Vimmer	Skandi (COCI20_skandi)	C++14	9 / 110	10041 ms	384 KiB

Main

#include <bits/stdc++.h>

//#pragma GCC optimize("unroll-loops")
//#pragma GCC optimize("-O3")
//#pragma GCC optimize("Ofast")
//#pragma GCC optimize("fast-math")
//#pragma GCC optimize("no-stack-protector")

#define F first
#define S second
#define sz(x) int(x.size())
#define pb push_back
#define N 75005
#define M ll(1e9 + 7)


using namespace std;

typedef long double ld;
typedef long long ll;
typedef short int si;






int main()
{
    ios_base::sync_with_stdio(0); istream::sync_with_stdio(0); cin.tie(0); cout.tie(0);

    int n, m;

    cin >> n >> m;

    string s[n];

    for (int i = 0; i < n; i++) cin >> s[i];

    vector <pair <int, int> > g; g.clear();

    for (int i = 0; i < n; i++)
      for (int j = 0; j < m; j++)
        if (s[i][j] == '1' && ((i + 1 != n && s[i + 1][j] == '0') || (j + 1 != m && s[i][j + 1] == '0'))) g.pb({i, j});

    int ans = 1e9;

    vector <pair <pair <int, int>, int> > gr; gr.clear();

    for (int umsk = 0; umsk < (1 << sz(g)); umsk++)
    {
        if (__builtin_popcount(umsk) > ans) continue;

        vector <int> pr; pr.clear();

        for (int i = 0; i < sz(g); i++) if ((1 << i) & umsk) pr.pb(i);

        for (int msk = 0; msk < (1 << sz(pr)); msk++)
        {
            bool gd = 1;

            bool mk[n][m];

            for (int i = 0; i < n; i++)
                for (int j = 0; j < m; j++) mk[i][j] = s[i][j] - '0';

            for (int i = 0; i < sz(pr); i++)
                if ((1 << i) & msk)
                {
                    int x = g[pr[i]].F, y = g[pr[i]].S;

                    if (y + 1 == m || s[x][y + 1] == '1') {gd = 0; break;}

                    int j = y + 1;

                    while (j < m && s[x][j] == '0') {mk[x][j] = 1; j++;}
                }
                else
                {
                    int x = g[pr[i]].F, y = g[pr[i]].S;

                    if (x + 1 == n || s[x + 1][y] == '1') {gd = 0; break;}

                    int j = x + 1;

                    while (j < n && s[j][y] == '0') {mk[j][y] = 1; j++;}
                }

            for (int i = 0; i < n; i++)
                for (int j = 0; j < m; j++) if (!mk[i][j]) gd = 0;

            if (gd)
            {
                ans = sz(pr);

                gr.clear();

                for (int i = 0; i < sz(pr); i++) gr.pb({{g[pr[i]].F, g[pr[i]].S}, (1 << i) & msk});

                break;
            }
        }
    }

    cout << sz(gr) << endl;

    for (auto it : gr) cout << it.F.F + 1 << " " << it.F.S + 1 << " " << (it.S == 1 ? "DESNO" : "DOLJE") << endl;
}

• Subtask 1: 9 / 18
• Subtask 2: 0 / 32
• Subtask 3: 0 / 60