제출 #1260980

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
1260980		Braabebo10	Red-blue table (IZhO19_stones)	C++20	53 / 100	815 ms	4884 KiB

stones

#include<bits/stdc++.h>
#define ll long long
#define nl "\n"
#define all(v) v.begin(),v.end()
#define baraa ios_base::sync_with_stdio(false);cin.tie(NULL);
using namespace std;

int main() {
    baraa
    auto solve = [&](ll n, ll m, vector<vector<char> > &res, ll &best) {
        for (ll a = max(0LL, n - 100); a <= n; a++) {
            vector<vector<char> > grid(n, vector<char>(m));
            vector<ll> neg[n];
            for (ll i = 0; i < n; i++)
                for (ll j = 0; j < m; j++)
                    grid[i][j] = (i < a and j <= m / 2 ? '+' : '-');
            ll b = 0;
            for (ll j = m - 1; j >= 0; j--) {
                ll delta = n / 2 + 1;
                for (ll i = 0; i < n; i++)
                    delta -= (grid[i][j] == '-');
                if (delta < 0) {
                    for (ll i = 0; i < abs(delta); i++)
                        neg[i].push_back(j);
                    b++;
                } else {
                    for (ll i = 0; i < n; i++) {
                        if (delta == 0)break;
                        if (neg[i].size()) {
                            delta--;
                            swap(grid[i][j], grid[i][neg[i].back()]);
                            neg[i].pop_back();
                        }
                    }
                }
                b += (delta == 0);
            }
            if (best < a + b) {
                best = a + b;
                res = grid;
            }
        }
    };
    ll t;
    cin >> t;
    while (t--) {
        ll n, m;
        cin >> n >> m;
        ll best1 = -1, best2 = -1;
        vector<vector<char> > g1(n, vector<char>(m)), g2(m, vector<char>(n));
        solve(n, m, g1, best1);
        solve(m, n, g2, best2);
        cout << max(best1, best2) << nl;
        if (best2 > best1) {
            for (ll i = 0; i < m; i++)
                for (ll j = 0; j < n; j++)
                    g1[j][i] = g2[i][j];
        }
        for (ll i = 0; i < n; i++) {
            for (ll j = 0; j < m; j++)
                cout << g1[i][j];
            cout << nl;
        }
    }
    return 0;
}

• Subtask 1: 17 / 17
• Subtask 2: 10 / 10
• Subtask 3: 0 / 16
• Subtask 4: 11 / 11
• Subtask 5: 15 / 15
• Subtask 6: 0 / 31