Submission #229107

#	Time	Username	Problem	Language	Result	Execution time	Memory
229107		kartel	Maja (COCI18_maja)	C++14	110 / 110	191 ms	764 KiB

maja

#include <bits/stdc++.h>
//#include <ext/pb_ds/assoc_container.hpp>
//#include <ext/pb_ds/tree_policy.hpp>
#define in(x) freopen(x, "r", stdin)
#define out(x) freopen(x, "w", stdout)
#pragma GCC optimize("Ofast")
#pragma GCC optimize("unroll-loops")
#pragma GCC optimize("-O3")
#define F first
#define S second
#define pb push_back
#define N +100500
#define M ll(1e9 + 7)
#define sz(x) (int)x.size()
#define re return
#define oo ll(1e18)
#define el '\n'
using namespace std;
//using namespace __gnu_pbds;
//typedef tree <int, null_type, less_equal <int> , rb_tree_tag, tree_order_statistics_node_update> ordered_set;
typedef long long ll;
typedef long double ld;

ll f[2][105][105];
ll a[105][105], i, j, q, nx, p, k, x, y, n, m, mx;

int main()
{
    srand(time(0));
    ios_base::sync_with_stdio(0);
    iostream::sync_with_stdio(0);
    ios::sync_with_stdio(0);
    cin.tie(NULL);
    cout.tie(NULL);

//    in("input.txt");
//    out("output.txt");

    cin >> n >> m >> x >> y >> k;

    k /= 2;

    for (i = 1; i <= n; i++)
        for (j = 1; j <= m; j++) cin >> a[i][j], f[0][i][j] = -1e18;

    mx = min(n * m, k);

    f[0][x][y] = 0;

    for (p = 0; p < mx; p++)
    {
        nx = (p + 1) % 2;
        q = p % 2;
        for (i = 1; i <= n; i++)
          for (j = 1; j <= m; j++) f[nx][i][j] = -1e18;

        for (i = 1; i <= n; i++)
          for (j = 1; j <= m; j++)
          {
              f[nx][i + 1][j] = max(f[nx][i + 1][j], f[q][i][j] + a[i + 1][j]);
              f[nx][i - 1][j] = max(f[nx][i - 1][j], f[q][i][j] + a[i - 1][j]);
              f[nx][i][j + 1] = max(f[nx][i][j + 1], f[q][i][j] + a[i][j + 1]);
              f[nx][i][j - 1] = max(f[nx][i][j - 1], f[q][i][j] + a[i][j - 1]);
          }
    }

    ll ans = 0;

    for (i = 1; i <= n; i++)
      for (j = 1; j <= m; j++)
    {
        int kl = k - mx;

        ll vl = 0;
        vl = max(vl, a[i - 1][j]);
        vl = max(vl, a[i + 1][j]);
        vl = max(vl, a[i][j - 1]);
        vl = max(vl, a[i][j + 1]);
        vl += a[i][j];

        ans = max(ans, f[mx % 2][i][j] * 2 + kl * vl - a[i][j]);
    }

    cout << ans;
}

• 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