1. Let the set $X=\{1,2, \cdots, 2017\}$,
$S=\{(x, y, z) \mid x, y, z \in X$, and the condition
$$
x<y<z, y<z<x, z<x<y
$$

holds for exactly one of them $\}$.
If $(x, y, z) \in S$, and $(z, w, x) \in S$, then the correct option is ( ).
(A) $(y, z, w) \in S$, and $(x, y, w) \notin S$
(B) $(y, z, w) \in S$, and $(x, y, w) \in S$
(C) $(y, z, w) \notin S$, and $(x, y, w) \in S$
(D) $(y, z, w) \notin S$, and $(x, y, w) \notin S$
- 1. B.

Given $(x, y, z) \in S$, we know
$$
x<y<z, \quad y<z<x, \quad z<x<y
$$

exactly one of these holds.
Given $(z, w, x) \in S$, we know
$$
z<w<x, \quad w<x<z, \quad x<z<w
$$

exactly one of these holds.
Without loss of generality, assume $x<y<z$, then $x<y<z<w$ or $w<x<y<z$.
In either case, we have
$$
(y, z, w) \in S \text {, and }(x, y, w) \in S \text {. }
$$