随便发发，不一定对，大部分是y总的板书，小部分是自己的理解

• 本题不能走两次摘花生，因为两次最优解不一定就是全局最优解，具体看视频下面的评论，有人提到了

C++代码

#include <iostream>
#include <algorithm>

using namespace std;

const int N = 15;

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

int main()
{
    cin >> n;
    int a, b, c;
    while (cin >> a >> b >> c, a || b || c) w[a][b] = c;

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

    cout << f[n * 2][n][n];

    return 0;
}