Submission #502632

#	Time	Username	Problem	Language	Result	Execution time	Memory
502632		stanislavpolyn	Red-blue table (IZhO19_stones)	C++17	100 / 100	1169 ms	2180 KiB

stones

#include <bits/stdc++.h>

#define fr(i, a, b) for(int i = (a); i <= (b); ++i)
#define rf(i, a, b) for(int i = (a); i >= (b); --i)
#define fe(x, y) for(auto& x : y)

#define fi first
#define se second
#define pb push_back
#define mp make_pair
#define mt make_tuple

#define all(x) (x).begin(), (x).end()
#define sz(x) (int)(x).size()
#define pw(x) (1LL << (x))

using namespace std;

#include <ext/pb_ds/assoc_container.hpp>
using namespace __gnu_pbds;
template<typename T>
using oset = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
#define fbo find_by_order
#define ook order_of_key

template<typename T>
bool umn(T& a, T b) { return (a > b ? (a = b, 1) : 0); }
template<typename T>
bool umx(T& a, T b) { return (a < b ? (a = b, 1) : 0); }

using ll = long long;
using ld = long double;
using pii = pair<int, int>;
using pll = pair<ll, ll>;
template<typename T>
using ve = vector<T>;

const int N = 1e3 + 5;

int n, m;
char a[N][N];
int cntR[N];
int cntB[N];

int solve(int n, int m, int R = 1e6) {
    fr(i, 1, n) cntR[i] = 0;
    fr(i, 1, m) cntB[i] = n;

    int needR = m / 2 + 1;
    int needB = n / 2 + 1;

    fr(i, 1, n) {
        fr(j, 1, m) {
            a[i][j] = '.';
        }
    }


    int have = m;

    int best = have;
    int bestR = 0;

    fr(i, 1, min(n, R)) {
        fr(T, 1, needR) {
            int mx = -1e9;
            int mxi = -1;
            fr(j, 1, m) {
                if(a[i][j] != 'R' && umx(mx, cntB[j])) {
                    mxi = j;
                }
            }
            cntB[mxi]--;
            if(cntB[mxi] < needB) have--;
            a[i][mxi] = 'R';
        }

        if(umx(best, i + have)) {
            bestR = i;
        }
    }

    fr(i, 1, n) {
        fr(j, 1, m) {
            if(a[i][j] != 'R') {
                a[i][j] = 'B';
            }
        }
    }

    fr(i, bestR + 1, n) {
        fr(j, 1, m) {
            a[i][j] = 'B';
        }
    }

    return best;
}

void solve() {
    cin >> n >> m;

    int score1 = 0;
    int score2 = 0;
    score1 = solve(n, m);
    score2 = solve(m, n);

    if(score1 > score2) {
        cout << solve(n, m) << "\n";
        fr(i, 1, n) {
            fr(j, 1, m) {
                cout << (a[i][j] == 'R' ? '+' : '-');
            }
            cout << "\n";
        }
        return;
    }

    cout << solve(m, n) << "\n";
    fr(i, 1, n) {
        fr(j, 1, m) {
            cout << (a[j][i] == 'R' ? '-' : '+');
        }
        cout << "\n";
    }
}

int main() {
#ifdef LOCAL
    freopen("input.txt", "r", stdin);
    freopen("output.txt", "w", stdout);
    ios::sync_with_stdio(0);
    cin.tie(0);
#else
//    freopen("input.txt", "r", stdin);
//    freopen("output.txt", "w", stdout);
    ios::sync_with_stdio(0);
    cin.tie(0);
#endif

//    int n, m;
//    cin >> n >> m;
//
////    cout << solve(n, m, n) << "\n";
//////
//    int mx = -1e9;
//    int mxi;
//    fr(i, 0, n) {
////        cout << solve(n, m, i) << " " << i << "\n";
//        if(umx(mx, solve(n, m, i))) {
//            mxi = i;
//        }
//    }
//    cout << mxi << "\n";
//
//    return 0;
//
//    fr(i, 0, n) {
//        cout << solve(n, m, i) << "\n";
//    }
//
//    return 0;

    int t;
    cin >> t;

    while(t--) {
        solve();
    }

    return 0;
}

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