我觉得这题跟 方格取数 这题没差

但我调半天都挑不出来气死了气死了气死了。

一定搞清楚各种定义！！！

比如：f[k][i1][i2]，$i1, i2$ 分别是是两个点的 $x$ 坐标，$k$ 应该是步数吧
所以最后输出的时候也应该写 f[n + m][n][n],而不是f[n + m][n][m]

(这个错怎么也找不出来，看来我应该锻炼人肉debug能力，并且要静下心来去想题，去想每一行是什么意思)
冷静思考！

#include<iostream>
#include<algorithm>

using namespace std;

const int N = 60;

int n, m;
int w[N][N];
int f[2 * N][N][N];

int main()
{
    scanf("%d%d", &n, &m);

    for (int i = 1; i <= n; i ++ )
            for (int j = 1; j <= m; j ++ )
                scanf("%d", &w[i][j]);

    for (int k = 2; k <= n + m; k ++ )
        for (int i1 = 1; i1 <= n; i1 ++ )
            for (int i2 = 1; i2 <= n; i2 ++ )
            {
                int j1 = k - i1, j2 = k - i2;
                if (j1 >= 1 && j1 <= m && j2 >= 1 && j2 <= m) 
                {
                    int t = w[i1][j1];
                    if (i1 != i2) t += w[i2][j2];
                    int &x = f[k][i1][i2];
                    x = max(x, f[k - 1][i1 - 1][i2 - 1] + t);
                    x = max(x, f[k - 1][i1 - 1][i2] + t);
                    x = max(x, f[k - 1][i1][i2 - 1] + t);
                    x = max(x, f[k - 1][i1][i2] + t);
                }
            }


    printf("%d\n", f[n + m][n][n]);

    return 0;
}

刚开始就觉得这俩题一样，于是抄上之后怎么也改不对，于是…

看了别人的写法，写暴力写出来了hh还挺爽

#include<iostream>
#include<algorithm>

using namespace std;

const int N = 60;

int n, m;
int w[N][N];
int f[N][N][N][N];

int main()
{
    scanf("%d%d", &n, &m);

    for (int i = 1; i <= n; i ++ )
            for (int j = 1; j <= m; j ++ )
                scanf("%d", &w[i][j]);

    for (int i1 = 1; i1 <= n ; i1 ++ )
        for (int j1 = 1; j1 <= m ; j1 ++ )
            for (int i2 = 1; i2 <= n ; i2 ++ )
                for (int j2 = 1; j2 <= m ; j2 ++ )
                {
                    int x = w[i1][j1] + w[i2][j2];
                    int a = max(f[i1-1][j1][i2-1][j2], f[i1-1][j1][i2][j2-1]);  // 下下   下右
                    int b = max(f[i1][j1-1][i2-1][j2], f[i1][j1-1][i2][j2-1]);  // 右下   右右

                    f[i1][j1][i2][j2] = max(a, b) + x;

                    if (i1 == i2 && j1 == j2) f[i1][j1][i2][j2] -= w[i1][j1];
                }

    printf("%d\n", f[n][m][n][m]);

    return 0;
}