One, (40 points) Integers $a, b, c, d$ satisfy $ad - bc = 1$. Find the minimum value of $a^2 + b^2 + c^2 + d^2 + ab + cd - ac - bd - bc$, and determine all quadruples $(a, b, c, d)$ that achieve this minimum value.
Let the required expression be $f(a, b, c, d)$. Then
$$
\begin{aligned}
f & (a, b, c, d) \\
= & \frac{1}{2}\left((a+b)^{2}+(c+d)^{2}+\right. \\
& \left.(a-c)^{2}+(b-d)^{2}\right)-b c .
\end{aligned}
$$
(1) If $b c>0$, then $a d=1+b c>0$.

Thus, $b$ and $c$ have the same sign, and $a$ and $d$ have the same sign.
Since $f(a, b, c, d)=f(-a,-b,-c,-d)$, we only need to consider the following two cases.
(i) $a, b, c, d>0$.
By $a+b \geqslant 1+b, c+d \geqslant c+1$, we have
$$
\begin{array}{l}
f(a, b, c, d) \\
\geqslant \frac{1}{2}\left((b+1)^{2}+(c+1)^{2}\right)-b c \\
=\frac{(b-c)^{2}}{2}+1+b+c>2 .
\end{array}
$$
(ii) $a, d>0, b, c<0$.
$$
\begin{array}{l}
f(a, b, c, d) \\
\geqslant \frac{1}{2}\left((a+b)^{2}+(c+d)^{2}+(a-c)^{2}+(b-d)^{2}\right)-b c \\
\geqslant \frac{1}{2}\left((a+b)^{2}+(a-c)^{2}+(c+d)^{2}+(b-d)^{2}\right)-b c \\
\geqslant \frac{1}{2}\left((1+b)^{2}+(1+c)^{2}+(1-c)^{2}+(1-b)^{2}\right)-b c \\
\geqslant \frac{1}{2}\left(4+2b+2c\right)-b c \\
\geqslant 2 .
\end{array}
$$
(2) If $b c=0$, then $a d=1$. Hence, $a=d= \pm 1$.
$$
\begin{array}{l}
\text { If } b=0, \text { then } \\
f(a, b, c, d) \\
=\frac{1}{2}\left(a^{2}+(c+d)^{2}+(a-c)^{2}+d^{2}\right) \\
=1+\frac{1}{2}\left((c-1)^{2}+(c+1)^{2}\right) \\
\geqslant 2,
\end{array}
$$

The equality holds if and only if $c=0$.
Similarly, if $c=0$, then $f(a, b, c, d) \geqslant 2$, the equality holds if and only if $b=0$.
(3) If $b c<0$, if $b c \leqslant-3$, then
$$
f(a, b, c, d) \geqslant 3 \text {. }
$$

If $b c=-2$, then $f(a, b, c, d) \geqslant 2$, the equality holds when $a=-b=c=-d$. But in this case, $b^{2}=2$ contradicts the fact that $b$ is an integer. Hence, the equality cannot be achieved.

If $b c=-1$, then $a d=0, f(a, b, c, d) \geqslant 1$, the equality holds when $a=-b=c=-d$. But $a d=0, b c=-1$, which is impossible.
Therefore, $f(a, b, c, d) \geqslant 2$, the equality holds when
$$
(a+b)^{2}+(a-c)^{2}+(c+d)^{2}+(b-d)^{2}=2 \text {. }
$$

By $b c=-1$, we know $b^{2}=c^{2}=1$.
By $a d=0$, we know that either $a$ or $d$ must be 0.
If $a=0$, then $b=-c=d$. Hence
$$
(a, b, c, d)=(0,1,-1,1) \text { or }(0,-1,1,-1) \text {. }
$$

Similarly, if $d=0$, then $a=-b=c$. Hence
$$
(a, b, c, d)=(1,-1,1,0) \text { or }(-1,1,-1,0) \text {. }
$$

In summary, the minimum value of $f(a, b, c, d)$ is 2, and the equality holds if and only if $(a, b, c, d)$ is one of $(0,1,-1,1),(0,-1,1,-1),(1,-1,1,0),(-1,1,-1,0)$, $(1,0,0,1),(-1,0,0,-1)$.