Submission #229165

#	Time	Username	Problem	Language	Result	Execution time	Memory
229165		Vimmer	Maja (COCI18_maja)	C++14	110 / 110	304 ms	760 KiB

maja

#include <bits/stdc++.h>

//#include <ext/pb_ds/assoc_container.hpp>
//#include <ext/pb_ds/tree_policy.hpp>

//#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) ll(x.size())
#define pb push_back
#define N 100005
#define MOD ll(998244353)

using namespace std;

//using namespace __gnu_pbds;

typedef long double ld;
typedef long long ll;

typedef short int si;

//typedef tree <int,null_type,less<int>,rb_tree_tag,tree_order_statistics_node_update> ordered_set;



ll f[2][105][105], ar[105][105], ans;

int main()
{
    //freopen("input.txt", "r", stdin);

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

    ll n, m, a, b, k;

    cin >> n >> m >> a >> b >> k;

    for (ll i = 1; i <= n; i++)
        for (ll j = 1; j <= m; j++) cin >> ar[i][j];

    int mn = min(n * m, k / 2);

    for (int i = 0; i <= n; i++)
        for (int j = 0; j <= m; j++) f[0][i][j] = f[1][i][j] = -1e18;

    f[0][a][b] = 0;

    for (int u = 0; u < mn; u++)
    {
        int nt = (u + 1) % 2, t = u % 2;

        for (int i = 1; i <= n; i++)
            for (int j = 1; j <= m; j++) f[nt][i][j] = -1e18;

        for (int i = 1; i <= n; i++)
            for (int j = 1; j <= m; j++) f[nt][i][j] = ar[i][j] + max(i - 1 == 0 ? -1e18 : f[t][i - 1][j], max(i + 1 > n ? -1e18 : f[t][i + 1][j], max(j - 1 == 0 ? -1e18 : f[t][i][j - 1], j + 1 > m ? -1e18 : f[t][i][j + 1])));
    }

    for (int i = 1; i <= n; i++)
        for (int j = 1; j <= m; j++)
        {
            ll sum = 2 * f[mn % 2][i][j];

            ll rst = k - mn * 2;

            ll mx = 0;

            if (i - 1 >= 1) mx = max(mx, ar[i - 1][j]);

            if (j - 1 >= 1) mx = max(mx, ar[i][j - 1]);

            if (i + 1 <= n) mx = max(mx, ar[i + 1][j]);

            if (j + 1 <= m) mx = max(mx, ar[i][j + 1]);

            sum += max(0ll, (rst / 2)) * mx;

            sum += max(0ll, ((rst / 2) - 1)) * ar[i][j];

            if (sum == 2 * f[mn % 2][i][j]) sum -= ar[i][j];

            ans = max(ans, sum);
        }

    cout << ans << endl;
}

• Subtask 1: 11 / 11
• Subtask 2: 11 / 11
• Subtask 3: 11 / 11
• Subtask 4: 11 / 11
• Subtask 5: 11 / 11
• Subtask 6: 11 / 11
• Subtask 7: 11 / 11
• Subtask 8: 11 / 11
• Subtask 9: 11 / 11
• Subtask 10: 11 / 11