Edge Groups

#树形结构 #组合数学 #树形dp

题目描述

Given an undirected connected graph of $n$ vertices and $n-1$ edges, where $n$ is guaranteed to be odd. You want to divide all the $n-1$ edges to $\frac{n-1}{2}$ groups under following constraints:

• There are exactly 2 edges in each group
• The 2 edges in the same group share a common vertex

Determine the number of valid dividing schemes modulo $998244353$. Two schemes are considered different if there are 2 edges that are in the same group in one scheme but not in the same group in the other scheme.

输入格式

The first line contains one integer $n(3\le n\le 10^5)$, denoting the number of vertices.

Following $n-1$ lines each contains two integers $u,v(1\le u\le v\le n)$, denoting that vertex $u,v$ are undirectedly connected by an edge.

It is guaranteed that $n$ is odd and that the given graph is connected.

输出格式

Output one line containing one integer, denoting the number of valid dividing schemes modulo $998244353$.

样例 #1

样例输入 #1

7
1 2
1 3
1 7
4 7
5 7
6 7

样例输出 #1

3

解题思路

观察发现，如果子树的边的个数为奇数的时候，需要和选一条边和父节点的连边组合，剩下的边可以两两组合。

如下图：

其中，$n$条边两两组合的方案数共有 ∏ i = 1 i ≤ n f a c t i , i ∈ { o d d } \prod_{i=1}^{i\leq n}fact_i \ , i \in \{odd\} ∏i=1i≤n​facti​ ,i∈{odd} 即不大于$n$的奇数的阶乘之积，可以通过数学归纳法证明，这里不多赘述。

因此，转移方程为：

f u = f u ∏ v ∏ i = 1 i ≤ n f v f a c t i ( v ∈ c h i l d u , i ∈ { o d d } ) f_u = f_u\prod_v \prod_{i=1}^{i \leq n} f_v fact_i \ ( v \in child_u,i \in \{ odd\}) fu​=fu​v∏​i=1∏i≤n​fv​facti​ (v∈childu​,i∈{odd})

代码

const int N = 1e6 + 7;vector<int> e[N];
int f[N], sz[N];void dfs(int u, int fa){f[u] = 1;for (auto &v : e[u]) {if (v == fa) continue;dfs(v, u);if (sz[v] % 2 == 0) sz[u]++;f[u] = f[u] * f[v] % mod;}if (sz[u] >= 2) {for (int i = 1; i <= sz[u]; i += 2) {f[u] = f[u] * i % mod;}}
}void solve()
{int n;cin >> n;for (int i = 1; i <= n - 1; i++) {int u, v;cin >> u >> v;e[u].push_back(v), e[v].push_back(u);}dfs(1, -1);cout << f[1]<< endl;
}signed main() {ios::sync_with_stdio(0);std::cin.tie(0);std::cout.tie(0);int t = 1;//cin >> t;while (t--) {solve();}
};