因为
1 <= i <= n, 1 <= k - i <= m
所以
i <= k - 1, i >= k - m;
max(1, k - m) <= i <= min(k - 1, n);

f[k][x1][x2] = max(f[k][x1][x2],    f[k - 1][x1][x2]            + t);
                                    f[k - 1][x1 - 1][x2]
                                    f[k - 1][x1][x2 - 1]
                                    f[k - 1][x1 - 1][x2 - 1]

所以可以写成a, b两层循环
for (int a = 0; a < 2; a++) {
    for (int b = 0; b < 2; b++) {
        f[k][x1][x2] = max(f[k][x1][x2], f[k - 1][x1 - a][x2 - b] + t);
    }
}

#include<iostream>
#include<algorithm>

using namespace std;
const int N = 55;
int f[N * 2][N][N], w[N][N];

int main() {
    int n, m;
    cin >> n >> m;
    for (int i = 1; i <= n; i++) {
        for (int j = 1; j <= m; j++) {
            cin >> w[i][j];
        }
    }

    for (int k = 2; k <= n + m; k++) {
        for (int x1 = max(1, k - m); x1 <= min(k - 1, n); x1++) {
            for (int x2 = max(1, k - m); x2 <= min(k - 1, n); x2++) {
                int t = w[x1][k - x1];
                if (x1 != x2) t += w[x2][k - x2];
                for (int a = 0; a < 2; a++) {
                    for (int b = 0; b < 2; b++) {
                        f[k][x1][x2] = max(f[k][x1][x2], f[k - 1][x1 - a][x2 - b] + t);
                    }
                }
            }
        }
    }

    cout << f[n + m][n][n] << endl;

}