In the first sample case there are $10$ possible tours, in order:

\begin{enumerate}
    \item from $0$ to $0$: the cost is $1$
    \item from $3$ to $3$: the cost is $1$
    \item from $1$ to $1$: the cost is $2$
    \item from $0$ to $1$: the cost is $1+2=3$
    \item from $2$ to $2$: the cost is $3$
    \item from $2$ to $3$: the cost is $3+1=4$
    \item from $1$ to $2$: the cost is $2+3=5$
    \item from $0$ to $2$: the cost is $1+2+3=6$
    \item from $1$ to $3$: the cost is $2+3+1=6$
    \item from $0$ to $3$: the cost is $1+2+3+1=7$
\end{enumerate}

from $0$ to $0$: the cost is $1$
from $3$ to $3$: the cost is $1$
from $1$ to $1$: the cost is $2$
from $0$ to $1$: the cost is $1+2=3$
from $2$ to $2$: the cost is $3$
from $2$ to $3$: the cost is $3+1=4$
from $1$ to $2$: the cost is $2+3=5$
from $0$ to $2$: the cost is $1+2+3=6$
from $1$ to $3$: the cost is $2+3+1=6$
from $0$ to $3$: the cost is $1+2+3+1=7$
 

The fourth tour starts from vineyard $0$ and end in vineyard $1$.