Submission #229014

#	Time	Username	Problem	Language	Result	Execution time	Memory
229014		NONAME	Maja (COCI18_maja)	C++17	44 / 110	2088 ms	640 KiB

maja

#include <bits/stdc++.h>
#define sz(x) int(x.size())
#define in(x) freopen(x, "r", stdin)
#define out(x) freopen(x, "w", stdout)
#define N 100500
#define oo ll(1e16)
#define ft first
#define sd second
#define pb push_back
#define ppb pop_back
#define el '\n'
#define elf endl
#define base ll(1e9 + 7)
using namespace std;
typedef long long ll;
typedef long double ld;

ll n, m, sx, sy, k, f[2][101][101], a[101][101], ans;

bool gd(int x, int y, int sx, int sy, int r) { return (abs(x - sx) + abs(y - sy) < r); }

int main() {
    ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);

//    in("input.txt");

    cin >> n >> m >> sx >> sy >> k;
    sx--; sy--;

    k /= 2;

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

    f[0][sx][sy] = 0;

    for (int t = 1; t <= k; t++) {
        int cr = t & 1;
        int lst = cr ^ 1;

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

        for (int i = 0; i < n; i++)
        for (int j = 0; j < m; j++) {
            if (f[lst][i][j] == -1)
                continue;

            if (i - 1 >= 0) f[cr][i - 1][j] = max(f[cr][i - 1][j], f[lst][i][j] + a[i - 1][j]);
            if (j - 1 >= 0) f[cr][i][j - 1] = max(f[cr][i][j - 1], f[lst][i][j] + a[i][j - 1]);
            if (i + 1 < n) f[cr][i + 1][j] = max(f[cr][i + 1][j], f[lst][i][j] + a[i + 1][j]);
            if (j + 1 < m) f[cr][i][j + 1] = max(f[cr][i][j + 1], f[lst][i][j] + a[i][j + 1]);
        }
    }

    int t = (k - 1) & 1;
    for (int i = 0; i < n; i++)
    for (int j = 0; j < m; j++) {
        if (!(abs(i - sx) + abs(j - sy) <= k))
            continue;
        if (i - 1 >= 0 && gd(i - 1, j, sx, sy, k)) ans = max(ans, f[t][i - 1][j] * 2 + a[i][j]);
        if (j - 1 >= 0 && gd(i, j - 1, sx, sy, k)) ans = max(ans, f[t][i][j - 1] * 2 + a[i][j]);
        if (i + 1 < n && gd(i + 1, j, sx, sy, k)) ans = max(ans, f[t][i + 1][j] * 2 + a[i][j]);
        if (j + 1 < m && gd(i, j + 1, sx, sy, k)) ans = max(ans, f[t][i][j + 1] * 2 + a[i][j]);
    }

    cout << ans;
}

• Subtask 1: 11 / 11
• Subtask 2: 11 / 11
• Subtask 3: 11 / 11
• Subtask 4: 11 / 11
• Subtask 5: 0 / 11
• Subtask 6: 0 / 11
• Subtask 7: 0 / 11
• Subtask 8: 0 / 11
• Subtask 9: 0 / 11
• Subtask 10: 0 / 11