牛客挑战赛30-T3 小G砍树

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题目大意：

n个节点的带标号无根树。每次选择一个度数为1的节点并将它从树上移除。问总共有多少种不同的方式能将这棵树删到只剩 1 个点。两种方式不同当且仅当至少有一步被删除的节点不同。

题解：

先考虑1号店最后移除时候的贡献，我们可以钦定1号点为根，并钦定他最后移除

然后就是一个树形dp

设\(f_i\)表示i号点子树移除方案数量，\(size_i\)表示1为根时子树大小

显然有dp式子\(f_x=\frac{(size_x-1)!}{\prod (size_i)!}\prod f_i\) (满足1为根时x是i的爹)

然后最后移除点1的情况的贡献就算出来了

我们考虑换根

先考虑x的转移少了j这个子树节点会咋样

那么\(f'_x=f_x/f_j/(size_x-1)!*(size_x-1-size_j)!*(size_j)!\)

注意这里的\(size_x\)是n，我们已经假设x是树的根了

然后我们再把\(f'_x\)转移到j上去，则新的j有

\(f'_j=f_j*f'_x/(size_j-1)!*(n-1)!/(n-size_i)!\)

由于涉及到求逆元，时间复杂度为\(O(n\log n)\)

#include <cstdio>
#include <vector>
using namespace std;

const int xkj = 998244353;

int n;
vector<int> out[100010];
int sz[100010], fa[100010], f[100010];
int fac[100010], inv[100010], ans;

int qpow(int x, int y)
{
	int res = 1;
	for (x %= xkj; y > 0; y >>= 1, x = x * (long long)x % xkj)
		if (y & 1) res = res * (long long)x % xkj;
	return res;
}

void dfs1(int x)
{
	sz[x] = 1, f[x] = 1;
	for (int i : out[x]) if (fa[x] != i)
	{
		fa[i] = x, dfs1(i), sz[x] += sz[i];
		f[x] = f[x] * (long long)f[i] % xkj;
		f[x] = f[x] * (long long)inv[sz[i]] % xkj;
	}
	f[x] = f[x] * (long long)fac[sz[x] - 1] % xkj;
}

void dfs2(int x)
{
	ans = (ans + f[x]) % xkj;
	for (int i : out[x]) if (fa[x] != i)
	{
		int tmp = f[x] * (long long)qpow(f[i], xkj - 2) % xkj * fac[sz[i]] % xkj * inv[n - 1] % xkj * fac[n - 1 - sz[i]] % xkj;
		int sb = f[i] * (long long)tmp % xkj * inv[sz[i] - 1] % xkj * fac[n - 1] % xkj * inv[n - sz[i]] % xkj; f[i] = sb;
		dfs2(i);
	}
}

int main()
{
	scanf("%d", &n);
	fac[0] = 1;
	for (int i = 1; i <= n; i++) fac[i] = fac[i - 1] * (long long)i % xkj;
	inv[n] = qpow(fac[n], xkj - 2);
	for (int i = n; i >= 1; i--) inv[i - 1] = inv[i] * (long long)i % xkj;
	for (int x, y, i = 1; i < n; i++) scanf("%d%d", &x, &y), out[x].push_back(y), out[y].push_back(x);
	dfs1(1), dfs2(1);
	printf("%d\n", ans);
	return 0;
}