DDOSvoid's Blog

loj 6289 花朵

发表于 分类于
本文字数： 8.9k 阅读时长 ≈ 8 分钟

题目描述

https://loj.ac/p/6289

简要题意：给定一个 $n$ 个点有点权的无根树，求所有大小为 $m$ 的点集的贡献和，要求点集内不存在任何两点相邻，一个点集的贡献是点集内所有点的点权积

$n\le 8\times 10^4$

Solution

我们令 $f_u(x)$ 表示选择 $u$ 的生成函数，$g_u(x)$ 表示不选 $u$ 的生成函数，容易得到转移 $f_u(x)=w_ux\prod_{v}g_v(x)$，$g_u(x)=\prod_{v}(f_v(x)+g_v(x))$

我们考虑轻重链剖分，轻儿子暴力合并，重链在链首合并，令 $G_u(x)=\prod_{v\neq son_u}g_v(x),F_u(x)=\prod_{v\neq son_u}(f_v(x)+g_v(x)) $，其中 $son_u$ 表示 $u$ 的重儿子，为了方便表示对于重链，我们不妨令其长度为 $k$，从链首到链尾的节点构成的序列为 $a_i$，那么我们有 $f_{a_i}(x)=w_uxg_{a_{i+1}}(x)G_{a_i}(x),g_{a_i}(x)=F_{a_i}(x)(f_{a_{i+1}}(x)+g_{a_{i+1}}(x))$，我们可以把转移写成矩阵的形式，然后分治计算即可

时间复杂度 $O(n\log^3 n)$

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
101
102
103
104
105
106
107
108
109
110
111
112
113
114
115
116
117
118
119
120
121
122
123
124
125
126
127
128
129
130
131
132
133
134
135
136
137
138
139
140
141
142
143
144
145
146
147
148
149
150
151
152
153
154
155
156
157
158
159
160
161
162
163
164
165
166
167
168
169
170
171
172
173
174
175
176
177
178
179
180
181
182
183
184
185
186
187
188
189
190
191
192
193
194
195
196
197
198
199
200
201
202
203
204
205
206
207
208
209
210
211
212
213
214
215
216
217
218
219
220
221
222
223
224
225
226
227
228
229
230
231
232
233
234
235
236
237
238
239
240
241
242
243
244
245
246
247
248
249
250
251
252
253
254
255
256
257
258
259
260
261
262
263
#include <iostream>
#include <vector>
#include <algorithm>
#include <array>
#define maxn 200010
#define ll long long
using namespace std;

const int p = 998244353;
inline int add(int x, int y) { return (x += y) >= p ? x - p : x; }
inline int mul(int x, int y) { return 1ll * x * y % p; }
inline int add(initializer_list<int> lst) { int s = 0; for (auto t : lst) s = add(s, t); return s; }
inline int mul(initializer_list<int> lst) { int s = 1; for (auto t : lst) s = mul(s, t); return s; }
int pow_mod(int x, int n) {
    int s = 1;
    for (; n; n >>= 1, x = mul(x, x))
        if (n & 1) s = mul(s, x);
    return s; 
}

#define Poly vector<int>
#define len(A) ((int) A.size())
namespace Pol {
    inline int add(int a, int b) { return (a += b) >= p ? a -= p : a; }
    inline int mul(int a, int b) { return 1ll * a * b % p; } 
    Poly operator - (const int &v, const Poly &a) {
        Poly res(a); 
        for (int i = 0; i < len(res); ++i) res[i] = p - res[i]; 
        res[0] = add(res[0], v); return res;
    }
    Poly operator - (const Poly &a, const int &v) {
        Poly res(a); res[0] = add(res[0], p - v); return res;
    }
    Poly operator * (const Poly &a, const int &v) {
        Poly res(a);
        for (int i = 0; i < len(res) ; ++i) res[i] = mul(res[i], v); 
        return res;
    }
    Poly operator + (const Poly &a, const Poly &b) {
        Poly res(a); if (len(res) < len(b)) res.resize(len(b));
        for (int i = 0; i < len(b); ++i) res[i] = add(res[i], b[i]);
        return res;
    }
 
    const int N = 4200000;
    const int G = 3;
    
    int P[N], inv[N], fac[N];
    void init_P(int n) {
        int l = 0; while ((1 << l) < n) ++l;
        for (int i = 0; i < n; ++i) P[i] = (P[i >> 1] >> 1) | ((i & 1) << l - 1);
    }
    void init_C() {
        if (fac[0]) return ;
        fac[0] = 1; for (int i = 1; i < N; ++i) fac[i] = mul(fac[i - 1], i);
        inv[N - 1] = pow_mod(fac[N - 1], p - 2); for (int i = N - 2; ~i; --i) inv[i] = mul(inv[i + 1], i + 1);
    }
    vector<int> init_W(int n) {
        vector<int> w(n); w[1] = 1;
        for (int i = 2; i < n; i <<= 1) {
            auto w0 = w.begin() + i / 2, w1 = w.begin() + i;
            int wn = pow_mod(G, (p - 1) / (i << 1));
            for (int j = 0; j < i; j += 2)
                w1[j] = w0[j >> 1], w1[j + 1] = mul(w1[j], wn);
        }
        return w; 
    } auto w = init_W(1 << 21);
    void DIT(Poly &a) {
        int n = len(a);
        for (int k = n >> 1; k; k >>= 1)
            for (int i = 0; i < n; i += k << 1)
                for (int j = 0; j < k; ++j) {
                    int x = a[i + j], y = a[i + j + k];
                    a[i + j + k] = mul(add(x, p - y), w[k + j]), a[i + j] = add(x, y);
                }
    }
    void DIF(Poly &a) {
        int n = len(a); 
        for (int k = 1; k < n; k <<= 1)
            for (int i = 0; i < n; i += k << 1)
                for (int j = 0; j < k; ++j) {
                    int x = a[i + j], y = mul(a[i + j + k], w[k + j]);
                    a[i + j + k] = add(x, p - y), a[i + j] = add(x, y);
                }
        int inv = pow_mod(n, p - 2);
        for (int i = 0; i < n; ++i) a[i] = mul(a[i], inv);
        reverse(a.begin() + 1, a.end());
    }
    Poly operator * (const Poly &A, const Poly &B) {
        int n = 1, n1 = len(A), n2 = len(B); while (n < n1 + n2 - 1) n <<= 1; init_P(n);
        Poly a(n), b(n);
        for (int i = 0; i < n1; ++i) a[i] = add(A[i], p);
        for (int i = 0; i < n2; ++i) b[i] = add(B[i], p);
        DIT(a); DIT(b);
        for (int i = 0; i < n; ++i) a[i] = mul(a[i], b[i]);
        DIF(a); a.resize(n1 + n2 - 1); return a; 
    }
    Poly MMul(const Poly &A, const Poly &B) { // 差卷积, 默认 A 和 B 的长度相同
        int n = 1, L = len(A); while (n < 2 * L - 1) n <<= 1; init_P(n);
        Poly a(n), b(n);
        for (int i = 0; i < L; ++i) a[i] = add(A[i], p);
        for (int i = 0; i < L; ++i) b[i] = add(B[i], p);
        reverse(b.begin(), b.begin() + L);
        DIT(a); DIT(b);
        for (int i = 0; i < n; ++i) a[i] = mul(a[i], b[i]);
        DIF(a); a.resize(L); reverse(a.begin(), a.end()); return a; 
    }
    Poly Der(const Poly &a) {
        Poly res(a);
        for (int i = 0; i < len(a) - 1; ++i) res[i] = mul(i + 1, res[i + 1]);
        res[len(a) - 1] = 0; return res;
    }
    Poly Int(const Poly &a) {
        static int inv[N];
        if (!inv[1]) {
            inv[1] = 1; 
            for (int i = 2; i < N; ++i) inv[i] = mul(p - p / i, inv[p % i]);
        }
        Poly res(a); res.resize(len(a) + 1);
        for (int i = len(a); i; --i) res[i] = mul(res[i - 1], inv[i]);
        res[0] = 0; return res;
    }
    Poly Inv(const Poly &a) {
        Poly res(1, pow_mod(a[0], p - 2)); 
        int n = 1; while (n < len(a)) n <<= 1; 
        for (int k = 2; k <= n; k <<= 1) {
            int L = 2 * k; init_P(L); Poly t(L);
            copy_n(a.begin(), min(k, len(a)), t.begin()); 
            t.resize(L); res.resize(L);
            DIT(res); DIT(t);
            for (int i = 0; i < L; ++i) res[i] = mul(res[i], add(2, p - mul(t[i], res[i])));
            DIF(res); res.resize(k);
        } res.resize(len(a)); return res;
    }
    Poly Offset(const Poly &a, int c) {
        int n = len(a); init_C();
        Poly t1(n), t2(n);
        for (int i = 0; i < n; ++i) t1[i] = mul(pow_mod(c, i), inv[i]);
        for (int i = 0; i < n; ++i) t2[i] = mul(a[i], fac[i]);
        t1 = MMul(t1, t2);
        for (int i = 0; i < n; ++i) t1[i] = mul(t1[i], inv[i]);
        return t1; 
    }
    pair<Poly, Poly> Divide(const Poly &a, const Poly &b) {
        int n = len(a), m = len(b); 
        Poly t1(a.rbegin(), a.rbegin() + n - m + 1), t2(b.rbegin(), b.rend()); t2.resize(n - m + 1);
        Poly Q = Inv(t2) * t1; Q.resize(n - m + 1); reverse(Q.begin(), Q.end());
        Poly R = Q * b; R.resize(m - 1); for (int i = 0; i < len(R); ++i) R[i] = add(a[i], p - R[i]);
        return make_pair(Q, R); 
    }
    Poly Ln(const Poly &a) {
        Poly res = Int(Der(a) * Inv(a));
        res.resize(len(a)); return res;
    }
    Poly Exp(const Poly &a) {
        Poly res(1, 1);
        int n = 1; while (n < len(a)) n <<= 1;
        for (int k = 2; k <= n; k <<= 1) {
            Poly t(res.begin(), res.end()); t.resize(k); t = Ln(t);
            for (int i = 0; i < min(len(a), k); ++i) t[i] = add(a[i], p - t[i]); t[0] = add(t[0], 1);
            res = res * t; res.resize(k);
        } res.resize(len(a)); return res;
    }
    Poly Sqrt(const Poly &a) { // a[0] = 1
        Poly res(1, 1); ll inv2 = pow_mod(2, p - 2); 
        int n = 1; while (n < len(a)) n <<= 1;
        for (int k = 2; k <= n; k <<= 1) {
            Poly t(res.begin(), res.end()), ta(a.begin(), a.begin() + min(len(a), k));
            t.resize(k); t = Inv(t) * ta; 
            res.resize(k); for (int i = 0; i < k; ++i) res[i] = mul(add(res[i], t[i]), inv2);
        } res.resize(len(a)); return res;
    }
    Poly Pow(const Poly &a, int k) { // a[0] = 1
        return Exp(Ln(a) * k);
    }
    Poly Pow(const Poly &a, int k, int kk) { 
        int n = len(a), t = n, m, v, inv, powv; Poly res(n);
        for (int i = n - 1; ~i; --i) if (a[i]) t = i, v = a[i];
        if (k && t >= (n + k - 1) / k) return res;
        if (t == n) { if (!k) res[0] = 1; return res; }
        m = n - t * k; res.resize(m);
        inv = pow_mod(v, p - 2); powv = pow_mod(v, kk); 
        for (int i = 0; i < m; ++i) res[i] = mul(a[i + (k > 0) * t], inv);
        res = Exp(Ln(res) * k); res.resize(n); 
        for (int i = m - 1; ~i; --i) {
            int tmp = mul(res[i], powv);
            res[i] = 0, res[i + t * k] = tmp;
        }
        return res;
    }
}  // namespace Pol
using Pol::operator*;
using Pol::operator+;

int n, m, w[maxn];

struct Edge {
    int to, next;
} e[maxn * 2]; int c1, head[maxn];
inline void add_edge(int u, int v) {
    e[c1].to = v; e[c1].next = head[u]; head[u] = c1++;
}

int sz[maxn], son[maxn], fa[maxn];
void pre(int u, int fa) {
    int Max = 0; sz[u] = 1; 
    for (int i = head[u]; ~i; i = e[i].next) {
        int v = e[i].to; if (v == fa) continue;
        pre(v, u); sz[u] += sz[v];
        if (sz[v] > Max) Max = sz[v], son[u] = v;
    } if (son[u]) ::fa[son[u]] = u; 
}

Poly Move(const Poly &a, int x) {
    Poly res(len(a) + 1);
    for (int i = 1; i <= len(a); ++i) res[i] = mul(a[i - 1], x);
    return res;
}

Poly merge(int l, int r, const vector<Poly> &vec) {
    if (l == r) return vec[l];
    int m = l + r >> 1;
    return Pol::operator*(merge(l, m, vec), merge(m + 1, r, vec)); 
}

Poly f[maxn], g[maxn], F[maxn], G[maxn];
array<Poly, 4> solve(int l, int r, const vector<int> &vec) {
    if (l == r)
        return { Poly{ 0 }, Move(G[vec[l]], w[vec[l]]), F[vec[l]], F[vec[l]] };
    int m = l + r >> 1; array<Poly, 4> ls = solve(l, m, vec), rs = solve(m + 1, r, vec);
    return { ls[0] * rs[0] + ls[1] * rs[2], ls[0] * rs[1] + ls[1] * rs[3],
            ls[2] * rs[0] + ls[3] * rs[2], ls[2] * rs[1] + ls[3] * rs[3] };
}

void dfs(int u, int fa) {
    vector<Poly> vf, vg; vector<int> vec;
    for (int i = head[u]; ~i; i = e[i].next) {
        int v = e[i].to; if (v == fa || v == son[u]) continue;
        dfs(v, u); vf.push_back(f[v]), vg.push_back(g[v]); 
    }
    if (vf.size()) {
        G[u] = merge(0, vf.size() - 1, vg);
        for (int i = 0; i < vf.size(); ++i) vf[i] = vf[i] + vg[i];
        F[u] = merge(0, vf.size() - 1, vf);
    } else F[u] = G[u] = Poly{ 1 };
    if (son[u]) dfs(son[u], u); if (::fa[u]) return ;
    int x = u; vec.push_back(u);
    while (son[x]) vec.push_back(x = son[x]); 
    array<Poly, 4> res = solve(0, vec.size() - 1, vec);
    f[u] = res[1], g[u] = res[3];
}

int main() { fill(head, head + maxn, -1); 
    ios::sync_with_stdio(false);
    cin.tie(nullptr); cout.tie(nullptr);

    cin >> n >> m;
    for (int i = 1; i <= n; ++i) cin >> w[i];
    for (int i = 1, x, y; i < n; ++i) cin >> x >> y, add_edge(x, y), add_edge(y, x);
    pre(1, 0); dfs(1, 0); Poly res = f[1] + g[1]; cout << (m < len(res) ? res[m] : 0) << "\n";
    return 0;
}