传送门
考虑枚举一条路径 (u,v),求出所有边经过它的答案
只需要求出 (u) 的子树内选出 (k) 个可以重复的点,使得它们到 (u) 的路径不相交
不难发现,就是从 (u) 的儿子的子树内各自选一个以及可以选多次 (u) 自己
设这个方案数为 (f_u)
再设 (size_u) 表示 (u) 的子树大小,(son_u) 表示 (u) 的儿子集合
考虑生成函数,设
[A(x)=prod_{vin son_u}(1+size_vx)
]
那么 (f_u=sum_{i=0}^{k}k^{underline i}A_i)
运用分治 (FFT) 预处理 (f) 可以做到 (O(nlog^2n))
现在的问题是当 (u,v) 是祖先关系的时候,设 (u) 为 (v) 的祖先
那么 (u) 要计算以 (v) 为根的时候的子树答案
一个想法就是两遍 (dfs) 每次到一条边 (u
ightarrow v) 对于这个多项式乘上一个
[frac{1+(n-size_u)x}{1+size_vx}
]
而考虑到一个点的儿子不同的 (size) 最多之后 (sqrt{n}) 个,那么只要做 (sqrt{n}) 次上面的操作,复杂度 (O(nsqrt{n}))
总复杂度 (O(nlog^2n+nsqrt{n}))
枚举路径可以通过子树和优化成 (O(n))
# include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxn(4e5 + 5);
const int mod(998244353);
inline void Inc(int &x, int y) {
x = x + y >= mod ? x + y - mod : x + y;
}
inline void Dec(int &x, int y) {
x = x - y < 0 ? x - y + mod : x - y;
}
inline int Add(int x, int y) {
return x + y >= mod ? x + y - mod : x + y;
}
inline int Sub(int x, int y) {
return x - y < 0 ? x - y + mod : x - y;
}
inline int Pow(ll x, int y) {
ll ret = 1;
for (; y; y >>= 1, x = x * x % mod)
if (y & 1) ret = ret * x % mod;
return ret;
}
int w[2][maxn], r[maxn], l, deg;
inline void Init(int len) {
int i, x, y;
for (l = 0, deg = 1; deg < len; deg <<= 1) ++l;
for (i = 0; i < deg; ++i) r[i] = (r[i >> 1] >> 1) | ((i & 1) << (l - 1));
x = Pow(3, (mod - 1) / deg), y = Pow(x, mod - 2), w[0][0] = w[1][0] = 1;
for (i = 1; i < deg; ++i) w[0][i] = (ll)w[0][i - 1] * x % mod, w[1][i] = (ll)w[1][i - 1] * y % mod;
}
inline void DFT(int *p, int opt) {
int i, j, k, t, wn, x, y;
for (i = 0; i < deg; ++i) if (r[i] < i) swap(p[r[i]], p[i]);
for (i = 1; i < deg; i <<= 1)
for (t = i << 1, j = 0; j < deg; j += t)
for (k = 0; k < i; ++k) {
wn = w[opt == -1][deg / t * k];
x = p[j + k], y = (ll)p[j + k + i] * wn % mod;
p[j + k] = Add(x, y), p[j + k + i] = Sub(x, y);
}
if (opt == -1) for (i = 0, wn = Pow(deg, mod - 2); i < deg; ++i) p[i] = (ll)p[i] * wn % mod;
}
int val[maxn], len, tmpa[20][maxn], tmpb[20][maxn], *a, *b;
void Solve(int l, int r, int *ret, int d) {
if (l == r) {
ret[0] = 1, ret[1] = val[l];
return;
}
int mid = (l + r) >> 1, tmp = r - l + 2, i;
Solve(l, mid, tmpa[d], d + 1), Solve(mid + 1, r, tmpb[d], d + 1);
a = tmpa[d], b = tmpb[d], Init(tmp), DFT(a, 1), DFT(b, 1);
for (i = 0; i < deg; ++i) a[i] = (ll)a[i] * b[i] % mod;
DFT(a, -1);
for (i = 0; i < tmp; ++i) ret[i] = a[i];
for (i = 0; i < deg; ++i) a[i] = b[i] = 0;
}
int n, k, first[maxn], cnt, size[maxn], ans, tmp[maxn], dv[maxn];
int f[maxn], fac[maxn], rec[maxn], inv[maxn], g[maxn], h[maxn];
struct Edge {
int to, next;
} edge[maxn];
inline void AddEdge(int u, int v) {
edge[cnt] = (Edge){v, first[u]}, first[u] = cnt++;
edge[cnt] = (Edge){u, first[v]}, first[v] = cnt++;
}
inline int Calc(int len) {
int i, ret = 0, r = min(len, k);
for (i = 0; i <= r; ++i) Inc(ret, (ll)tmp[i] * fac[k] % mod * inv[k - i] % mod);
return ret;
}
inline void PolyDiv(int len, int v) {
int i, inv, invv = Pow(v, mod - 2);
for (i = 0; i <= len; ++i) dv[i] = tmp[i], tmp[i] = 0;
for (i = len; i; --i)
if (dv[i]) {
tmp[i - 1] = inv = (ll)dv[i] * invv % mod;
Dec(dv[i], (ll)inv * invv % mod), Dec(dv[i - 1], inv);
}
}
inline void PolyMul(int len, int v) {
int i;
for (i = len; i; --i) Inc(tmp[i], (ll)tmp[i - 1] * v % mod);
}
inline int Cmp(int x, int y) {
return size[x] < size[y];
}
void Dfs1(int u, int ff) {
int e, v, i, j;
size[u] = 1;
for (e = first[u]; ~e; e = edge[e].next)
if ((v = edge[e].to) ^ ff) Dfs1(v, u), Inc(g[u], g[v]), size[u] += size[v];
len = 0;
for (e = first[u]; ~e; e = edge[e].next)
if ((v = edge[e].to) ^ ff) val[++len] = size[v];
if (!len) f[u] = 1, Inc(g[u], 1);
else {
Solve(1, len, tmp, 0), f[u] = Calc(len), Inc(g[u], f[u]), len = 0;
for (e = first[u]; ~e; e = edge[e].next)
if ((v = edge[e].to) ^ ff) val[++len] = v;
sort(val + 1, val + len + 1, Cmp);
for (i = 0; i <= len; ++i) rec[i] = tmp[i];
for (i = 1; i <= len; ++i) {
if (size[val[i]] == size[val[i - 1]]) h[val[i]] = h[val[i - 1]];
else {
for (j = 0; j <= len; ++j) tmp[j] = rec[j];
PolyDiv(len, size[val[i]]), PolyMul(len, n - size[u]);
h[val[i]] = Calc(len);
}
}
}
}
void Dfs2(int u, int ff) {
int e, v;
for (e = first[u]; ~e; e = edge[e].next)
if ((v = edge[e].to) ^ ff) Inc(ans, (ll)Sub(h[v], f[u]) * g[v] % mod), Dfs2(v, u);
}
int main() {
memset(first, -1, sizeof(first));
int i, u, v, sum = 0, mx;
scanf("%d%d", &n, &k);
if (k == 1) return printf("%lld
", ((ll)n * (n - 1) >> 1) % mod), 0;
mx = max(n, k);
for (i = 1; i < n; ++i) scanf("%d%d", &u, &v), AddEdge(u, v);
inv[0] = inv[1] = 1;
for (i = 2; i <= mx; ++i) inv[i] = (ll)(mod - mod / i) * inv[mod % i] % mod;
for (i = 2; i <= mx; ++i) inv[i] = (ll)inv[i - 1] * inv[i] % mod;
for (i = fac[0] = 1; i <= mx; ++i) fac[i] = (ll)fac[i - 1] * i % mod;
Dfs1(1, 0), Dfs2(1, 0);
for (i = 1; i <= n; ++i) Inc(sum, f[i]);
sum = (ll)sum * sum % mod;
for (i = 1; i <= n; ++i) Dec(sum, (ll)f[i] * f[i] % mod);
sum = (ll)sum * inv[2] % mod;
Inc(ans, sum), printf("%d
", ans);
return 0;
}