7.113 The complete works of Tolstoy consist of 100 volumes, which are randomly arranged on a bookshelf. Each time, it is allowed to swap the positions of any two volumes with different parity numbers. How many such swaps are required at a minimum to ensure that they are arranged in order by volume number?
[Solution] Use 100 points $x_{1}, x_{2}, \cdots, x_{100}$ to represent 100 positions on a bookshelf. When the $i$-th position holds the $j$-th volume of a book $(i \neq j)$, draw a directed edge from $x_{i}$ to $x_{j}$. After all such edges are drawn, we obtain a directed graph where each vertex has a degree of 2 or 0, and for vertices with degree 2, exactly one edge points to it and one edge leaves it. Note that a vertex $x_{i}$ with degree 0 means that the $i$-th position already holds the $i$-th volume of the book.

Removing these isolated vertices with degree 0, we get a directed graph where each vertex has a degree of 2. By a graph theory theorem, it can be decomposed into several disjoint cycles. When two books are swapped, if the two books are in different cycles, the swap will merge the two cycles into one, which we call "joining"; if the two books are in the same cycle and the corresponding two points are adjacent, one cycle will split into two (or possibly degenerate into an isolated point), which we call "splitting".

Now assume that among the cycles with length greater than 1 (i.e., at least two vertices), there are $a$ cycles consisting only of even-numbered vertices, and $b$ cycles consisting only of odd-numbered vertices. Clearly, $a \leqslant 25, b \leqslant 25$. Without loss of generality, let $0 < b \leqslant a$. Thus, after at most $a$ "joinings", all cycles will contain both odd-numbered and even-numbered vertices. At this point, only "splitting" is needed. In fact, each split can be performed on a pair of adjacent vertices, making at least one of them an isolated vertex, and when there are more odd (even) numbered vertices in the cycle, the odd (even) numbered vertex is made an isolated vertex. Thus, the splitting can continue until all cycles are isolated points. Clearly, such splitting requires at most 99 times. Therefore, the entire exchange process requires at most 124 exchanges.

On the other hand, when 50 even-numbered vertices form 1 cycle and 50 odd-numbered vertices form 25 cycles, to eliminate the odd-numbered cycles, at least 25 joining exchanges are needed. Additionally, suppose that throughout the exchange process, there are $k$ cycles that contain both even and odd vertices, then the only cycle with even vertices must first undergo $k-1$ splits. Let the number of vertices in these $k$ cycles be $n_{1}, n_{2}, \cdots, n_{k}$. Since a cycle with $n_{j}$ vertices needs at least $n_{j}-1$ splits to become $n_{j}$ isolated points, the total number of splits required is at least
$$
\left(n_{1}-1\right)+\left(n_{2}-1\right)+\cdots+\left(n_{k}-1\right)+(k-1)=99 \text{. }
$$

This indicates that at least $25+99=124$ exchanges are needed.
In summary, the minimum number of exchanges required is 124.