Submission #229065

#	Time	Username	Problem	Language	Result	Execution time	Memory
229065		ne4eHbKa	Maja (COCI18_maja)	C++17	110 / 110	380 ms	760 KiB

maja

//{ <defines>
#include <bits/stdc++.h>
using namespace std;

//#pragma comment(linker, "/stack:200000000")
//#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#pragma GCC optimize("Ofast,no-stack-protector,unroll-loops,fast-math,-O3")

#define fr(i, n) for(int i = 0; i < n; ++i)
#define fo(n) fr(i, n)

#define re return
#define ef else if
#define ifn(x) if(!(x))
#define _  << ' ' <<

#define ft first
#define sd second
#define ve vector
#define pb push_back
#define eb emplace_back

#define sz(x) int((x).size())
#define ip2(x) (1 << (x))
#define lp2(x) (1ll << (x))
#define bnd(x) x.begin(), x.end()
#define clr(x, y) memset((x), (y), sizeof (x))

typedef long long ll;
typedef long double ld;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
typedef ve<int> vi;

inline ll time() {re chrono :: system_clock().now().time_since_epoch().count();}
mt19937 rnd(time());
mt19937_64 RND(time());

template<typename t> inline void umin(t &a, t b) {a = min(a, b);}
template<typename t> inline void umax(t &a, t b) {a = max(a, b);}

int md = 998244353;

inline int m_add(int&a, int b) {a += b; if(a < 0) a += md; if(a >= md) a -= md; re a;}
inline int m_sum(int a, int b) {a += b; if(a < 0) a += md; if(a >= md) a -= md; re a;}
inline int m_mul(int&a, int b) {re a = 1ll * a * b % md;}
inline int m_prod(int a, int b) {re 1ll * a * b % md;}

int m_bpow(ll A, ll b) {
    int a = A % md;
    ll ans = 1;
    for(ll p = lp2(63 - __builtin_clzll(b)); p; p >>= 1) {
        (ans *= ans) %= md;
        if(p & b)
            (ans *= a) %= md;
    }
    re ans;
}

//const ld pi = arg(complex<ld>(-1, 0));
//const ld pi2 = pi + pi;
const int oo = 2e9;
const ll OO = 4e18;
//} </defines>
const int N = 105;

int n, m, a, b, k;
ll c[N][N];

void solve() {
    cin >> n >> m >> a >> b >> k;
    --a, --b;
    k >>= 1;
    fo(n) fr(j, m) cin >> c[i][j];
    int mx = min(n * m, k);
    ve< ve<ll> > f, g;
    f.assign(n, ve<ll>(m, -1));
    g = f;
    f[a][b] = 0;
    for(int t = 0; t < mx; ++t) {
        for(int i = 0; i < n; ++i) {
            for(int j = 0; j < m; ++j) {
                if(i > 0 && ~f[i - 1][j])
                    umax(g[i][j], f[i - 1][j] + c[i][j]);
                if(i < n - 1 && ~f[i + 1][j])
                    umax(g[i][j], f[i + 1][j] + c[i][j]);
                if(j > 0 && ~f[i][j - 1])
                    umax(g[i][j], f[i][j - 1] + c[i][j]);
                if(j < m - 1 && ~f[i][j + 1])
                    umax(g[i][j], f[i][j + 1] + c[i][j]);
            }
        }
        swap(f, g);
    }
    ll ans = 0;
    fr(i, n) fr(j, m) {
        int t = k - mx;
        if(t < 0) continue;
        ll v = 0;
        if(i) umax(v, c[i - 1][j]);
        if(j) umax(v, c[i][j - 1]);
        if(i + 1 < n) umax(v, c[i + 1][j]);
        if(j + 1 < m) umax(v, c[i][j + 1]);
        v += c[i][j];
        umax(ans, f[i][j] * 2 + v * t - c[i][j]);
    }
    cout << ans << endl;
}

int main() {
#ifdef _LOCAL
    freopen("in.txt", "r", stdin);
    int tests; cin >> tests;
    for(int test = 1; test <= tests; ++test) {
        cerr << test << " {\n";
        solve();
        cerr << "}\n\n";
    }
#else
    //freopen("input.txt", "r", stdin); freopen("output.txt", "w", stdout);
    ios_base :: sync_with_stdio(0); cin.tie(0); cout.tie(0);
    solve();
#endif
    return 0;
}

• 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