【LOJ】#3102. 「JSOI2019」神經網絡

LOJ#3102. 「JSOI2019」神經網絡

首先咱們容易發現就是把樹拆成若干條鏈，而後要求這些鏈排在一個環上，同一棵樹的鏈不相鄰

把樹拆成鏈能夠用一個簡單（可是須要複雜的分類討論）的樹揹包實現

\(dp[u][j][0/1/2]\)表示第\(u\)個點已經選了\(j\)條鏈，0是兩個不一樣子樹的鏈拼到一塊兒，1是隻有1個點，2是有一條至少有兩個點的鏈

經過這個咱們能夠求一個\(f[k]\)表示把這棵樹分紅\(k\)條鏈有幾種狀況

環排列能夠經過全排列除以排列長度獲得

咱們設把\(k\)條鏈分紅\(h\)個小塊，這樣咱們至少有了\(k - h\)對點同一棵樹且相鄰，容斥係數乘上\((-1)^{k - h}\)，對於全排列來講，咱們還須要除以\(h!\)

因此列出一個這樣的EGF

\(f[k]k!\binom{k - 1}{h - 1} \frac{x^{h}}{h!}\)

就能夠卷積了

而後對於答案的\(x^{h}\)乘上\((h - 1)!\)由於是環排

就能夠獲得答案

#include <bits/stdc++.h>
#define fi first
#define se second
#define pii pair<int,int>
#define mp make_pair
#define pb push_back
#define space putchar(' ')
#define enter putchar('\n')
#define eps 1e-10
#define ba 47
#define MAXN 5005
//#define ivorysi
using namespace std;
typedef long long int64;
typedef unsigned int u32;
typedef double db;
template<class T>
void read(T &res) {
    res = 0;T f = 1;char c = getchar();
    while(c < '0' || c > '9') {
    if(c == '-') f = -1;
    c = getchar();
    }
    while(c >= '0' && c <= '9') {
    res = res * 10 +c - '0';
    c = getchar();
    }
    res *= f;
}
template<class T>
void out(T x) {
    if(x < 0) {x = -x;putchar('-');}
    if(x >= 10) {
    out(x / 10);
    }
    putchar('0' + x % 10);
}
const int MOD = 998244353,MAXL = (1 << 15);
int W[MAXL + 5];
int fac[100005],invfac[100005];
int inc(int a,int b) {
    return a + b >= MOD ? a + b - MOD : a + b;
}
int mul(int a,int b) {
    return 1LL * a * b % MOD;
}
void update(int &x,int y) {
    x = inc(x,y);
}
int fpow(int x,int c) {
    int res = 1,t = x;
    while(c) {
    if(c & 1) res = mul(res,t);
    t = mul(t,t);
    c >>= 1;
    }
    return res;
}
int C(int n,int m) {
    if(n < m) return 0;
    else return mul(fac[n],mul(invfac[m],invfac[n - m]));
}
void NTT(vector<int> &f,int l,int on) {
    f.resize(l);
    for(int i = 1,j = l >> 1 ; i < l - 1 ; ++i) {
    if(i < j) swap(f[i],f[j]);
    int k = l >> 1;
    while(j >= k) {
        j -= k;
        k >>= 1;
    }
    j += k;
    }
    for(int h = 2 ; h <= l ; h <<= 1) {
    int wn = W[(MAXL + on * MAXL / h) % MAXL];
    for(int k = 0 ; k < l ; k += h) {
        int w = 1;
        for(int j = k ; j < k + h / 2 ; ++j) {
        int u = f[j],t = mul(w,f[j + h / 2]);
        f[j] = inc(u,t);
        f[j + h / 2] = inc(u,MOD - t);
        w = mul(w,wn);
        }
    }
    }
    if(on == -1) {
    int invL = fpow(l,MOD - 2);
    for(int i = 0 ; i < l ; ++i) f[i] = mul(f[i],invL);
    }
}
vector<int> operator * (vector<int> a,vector<int> b) {
    vector<int> c;
    int l = 1;
    while(l <= a.size() - 1 + b.size() - 1) l <<= 1;
    NTT(a,l,1);NTT(b,l,1);
    c.resize(l);
    for(int i = 0 ; i < l ; ++i) c[i] = mul(a[i],b[i]);
    NTT(c,l,-1);
    int s = c.size() - 1;
    while(s > 0) {
    if(c[s] == 0) {c.pop_back();--s;}
    else break;
    }
    return c;
}
void Init() {
    W[0] = 1;W[1] = fpow(3,(MOD - 1) / MAXL); 
    for(int i = 2 ; i < MAXL ; ++i) W[i] = mul(W[i - 1],W[1]);
    fac[0] = 1;
    for(int i = 1 ; i <= 100000 ; ++i) fac[i] = mul(fac[i - 1],i);
    invfac[100000] = fpow(fac[100000],MOD - 2);
    for(int i = 99999 ; i >= 0 ; --i) invfac[i] = mul(invfac[i + 1],i + 1);
}
int M,K;
struct node {
    int to,next;
}E[100005];
int head[5005],sumE,siz[5005];
int dp[5005][5005][3],g[5005][3],f[5005],all;
vector<int> z,ans;
void add(int u,int v) {
    E[++sumE].to = v;
    E[sumE].next = head[u];
    head[u] = sumE;
}
void dfs(int u,int fa) {
    dp[u][0][1] = 1;
    siz[u] = 1;
    for(int i = head[u] ; i ; i = E[i].next) {
    int v = E[i].to;
    if(v != fa) {
        dfs(v,u);
        for(int j = 0 ; j <= siz[u] + siz[v] ; ++j) memset(g[j],0,sizeof(g[j]));
        for(int j = 0 ; j <= siz[u] ; ++j) {
        for(int h = 0 ; h <= siz[v] ; ++h) {
            int t0 = inc(dp[v][h][1],mul(dp[v][h][2],2));
            int t1 = inc(dp[v][h][1],dp[v][h][2]);
            update(g[j + h][0],mul(dp[u][j][0],dp[v][h][0]));
            update(g[j + h + 1][0],mul(dp[u][j][0],t0));;
            update(g[j + h][1],mul(dp[u][j][1],dp[v][h][0]));
            update(g[j + h + 1][1],mul(dp[u][j][1],t0));
            update(g[j + h][2],mul(dp[u][j][1],t1));
            update(g[j + h][2],mul(dp[u][j][2],dp[v][h][0]));
            update(g[j + h + 1][2],mul(dp[u][j][2],t0));
            update(g[j + h + 1][0],mul(dp[u][j][2],mul(2,t1)));
        }
        }
        siz[u] += siz[v];
        for(int j = 0 ; j <= siz[u] ; ++j) {
        for(int h = 0 ; h < 3 ; ++h) {
            dp[u][j][h] = g[j][h];
        }
        }
    }
    }
    if(!fa) {
    memset(f,0,sizeof(f));
    for(int j = 0 ; j <= siz[1] ; ++j) {
        update(f[j],dp[1][j][0]);
        update(f[j + 1],dp[1][j][1]);
        update(f[j + 1],mul(2,dp[1][j][2]));
    }
    }
}
void Solve() {
    read(M);
    ans.pb(1);
    for(int i = 1 ; i <= M ; ++i) {
    sumE = 0;
    memset(head,0,sizeof(head));
    memset(siz,0,sizeof(siz));
    for(int j = 0 ; j <= K ; ++j) {
        for(int h = 0 ; h <= K ; ++h) {
        memset(dp[j][h],0,sizeof(dp[j][h]));
        }
    }
    read(K);
    all += K;
    int a,b;
    for(int j = 1 ; j < K ; ++j) {
        read(a);read(b);
        add(a,b);add(b,a);
    }
    dfs(1,0);
    z.clear();
    z.resize(K + 1);
    for(int j = K ; j >= 1 ; --j) {
        int a = mul(f[j],fac[j]);
        for(int h = j ; h >= 1 ; --h) {
        int t = mul(a,C(j - 1,h - 1));
        if((j - h) & 1) t = MOD - t;
        update(z[h],t);
        }
    }
    for(int j = 0 ; j <= K ; ++j) z[j] = mul(z[j],invfac[j]);
    ans = ans * z;
    }
    int res = 0;
    for(int i = 1 ; i <= all ; ++i) {
    update(res,mul(ans[i],fac[i - 1]));
    }
    out(res);enter;
}
int main(){
#ifdef ivorysi
    freopen("f1.in","r",stdin);
#endif
    Init();
    Solve();
}