【LOJ】#2983. 「WC2019」數樹

LOJ2983. 「WC2019」數樹

task0

有$i$條邊同樣答案就是$y^{n - i}$

task1

這裏有個避免容斥的方法，若是有$i$條邊重複咱們要算的是$y^{n - i}$，設$a = y^{-1}$那麼咱們能夠對於選了i條邊的方案算$a^{i}$

但是這樣須要容斥，因此有個神奇的技巧

$(a - 1 + 1)^{i} = \sum_{j = 0}^{i}(a - 1)^{j}\binom{i}{j}$

這樣，對於至少選了$j$條邊的方案，每選一條邊乘上一個$(a - 1)$就能夠避免容斥了（如下設$z = a - 1$)

分紅$m$個聯通塊方案數是

$n^{m - 2}\prod_{i = 1}^{m}a_{i}$

這至關於斷開一條邊乘一下N，且每一個聯通塊要選出一個點，dp完了以後答案再乘上$N^{-1}$

就直接設$dp[i][0/1]$表示$i$所在的聯通塊是否選了一個點的方案數

task2

仍是上面那個技巧，而後要爆推式子，假如硬點$i$條邊被選了的有$f_{i}$棵樹

$\sum_{i = 1}^{N - 1}f_iz^{i}$ $$ f_{i} = \frac{N!}{\prod_{k = 1}^{M}a_{k}!} \frac{1}{(N - x)!} \prod_{i = k}^{M}a_{k}^{a_{k} - 2} (n^{n - x - 2} \prod_{k = 1}^{M}a_{k})^{2} $$

從左到右分別是

給每一個聯通塊分配點，消除聯通塊的順序，聯通塊內生成樹的個數，聯通塊造成的樹的個數

有點亂，給合一下就是 $$ f_{i} = N! \prod_{k = 1}^{M}\frac{a_{k}^{a_{k}}}{a_{k}!} \frac{n^{2(n - x - 2)}}{(N - x)!} $$

而後咱們把答案帶進去 $$ ans = N!\sum_{i = 1}^{N} \frac{Z^{N - i}n^{2i - 4}}{i!}\prod_{k = 1}^{M}\frac{a_{k}^{a_{k}}}{a_{k}!} $$ 咱們給$a_{k}$設個生成函數，去掉和i無關的項就是 $$ ans = N!\frac{Z^{N}}{n^{4}}\sum_{i = 1}^{N}\frac{\frac{n^{2i}}{Z^{i}}\prod_{k = 1}^{M}\frac{a_{k}^{a_{k}}x^{a_{k}}}{a_{k}!} }{i!} $$

由於$M = i$

因此也能夠寫成 $$ ans = N!\frac{Z^{N}}{n^{4}}\sum_{i = 1}^{N}\frac{x^{n}^{i} }{i!} $$

因而設$F(x) = \sum_{i = 1}^{\infty} \frac{n^{2}}{Z}\frac{i^{i}}{i!}x^{i}$

很容易知道上面後半部分的式子是$e^{F(x)}$，而後乘上前面的係數就作完了

#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 MAXN 100005
#define ba 47
//#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 << 20);
int N,Y,op,W[MAXL + 5];
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;
}
namespace task0 {
    bool vis[MAXN];
    map<pii,int> zz;
    void Main() {
	int a,b;
	for(int i = 1 ; i < N ; ++i) {
	    read(a);read(b);
	    if(a > b) swap(a,b);
	    zz[mp(a,b)] = 1;
	}
	int ans = fpow(Y,N);
	int t = fpow(Y,MOD - 2);
	for(int i = 1 ; i < N ; ++i) {
	    read(a);read(b);
	    if(a > b) swap(a,b);
	    if(zz[mp(a,b)]) ans = mul(ans,t);
	}
	out(ans);enter;
    }
}
namespace task1 {
    struct node {
	int to,next;
    }E[MAXN * 2];
    int head[MAXN],sumE,Z;
    int dp[MAXN][2],tmp[2];
    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,dp[u][1] = 1;
	for(int i = head[u] ; i ; i = E[i].next) {
	    int v = E[i].to;
	    if(v != fa) {
		dfs(v,u);
		tmp[0] = tmp[1] = 0;
		update(tmp[1],mul(mul(dp[u][1],dp[v][1]),N));
		update(tmp[1],mul(mul(dp[u][1],dp[v][0]),Z));
		update(tmp[0],mul(mul(dp[u][0],dp[v][0]),Z));
		update(tmp[0],mul(mul(dp[u][0],dp[v][1]),N));
		update(tmp[1],mul(mul(dp[u][0],dp[v][1]),Z));
		dp[u][0] = tmp[0];dp[u][1] = tmp[1];
	    }
	}
    }
    void Main() {
	int a,b;
	for(int i = 1 ; i < N ; ++i) {read(a);read(b);add(a,b);add(b,a);}
	Z = fpow(Y,MOD - 2);Z = inc(Z,MOD - 1);
	dfs(1,0);
	int ans = dp[1][1];
	ans = mul(ans,fpow(N,MOD - 2));ans = mul(ans,fpow(Y,N));
	out(ans);enter;
    }
}
namespace task2 {
    int fac[MAXN],invfac[MAXN],inv[MAXN];
    vector<int> f,g;
    void NTT(vector<int> &p,int L,int on) {
	p.resize(L);
	for(int i = 1,j = L >> 1; i < L - 1 ; ++i) {
	    if(i < j) swap(p[i],p[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 = p[j],t = mul(w,p[j + h / 2]);
		    p[j] = inc(u,t);
		    p[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) p[i] = mul(p[i],invL);
	}
    }
    
    void limit(vector<int> &p,int N) {
	if(p.size() > N) p.resize(N);
    }
    vector<int> operator * (vector<int> a,vector<int> b) {
	int t = a.size() + b.size() - 2;
	int L = 1;
	while(L <= t) L <<= 1;
	NTT(a,L,1);NTT(b,L,1);
	vector<int> c;c.resize(L);
	for(int i = 0 ; i < L ; ++i) c[i] = mul(a[i],b[i]);
	NTT(c,L,-1);
	return c;
    }
    vector<int> Derivative(vector<int> p) {
	vector<int> res(p.size() - 1);
	for(int i = 0 ; i < p.size() - 1 ; ++i) {
	    res[i] = mul(p[i + 1],i + 1);
	}
	if(res.size() == 0) res.pb(0);
	return res;
    }
    vector<int> Integral(vector<int> p) {
	vector<int> res(p.size() + 1);
	for(int i = 1 ; i <= p.size() ; ++i) {
	    res[i] = mul(p[i - 1],inv[i]);
	}
	return res;
    }
    vector<int> inverse(vector<int> p,int len) {
	vector<int> g;
	if(len == 1) {
	    g.pb(fpow(p[0],MOD - 2));
	    return g;
	}
	g = inverse(p,len >> 1);
	int t = p.size() - 1 + 2 * (g.size() - 1);
	int L = 1;
	while(L <= t) L <<= 1;
	NTT(p,L,1);NTT(g,L,1);
	for(int i = 0 ; i < L ; ++i) {
	    g[i] = inc(mul(2,g[i]),MOD - mul(mul(g[i],g[i]),p[i]));
	}
	NTT(g,L,-1);
	g.resize(len);
	return g;
    }
    vector<int> Inverse(vector<int> p) {
	int L = 1,t = p.size() - 1;
	while(L <= t) L <<= 1;
	return inverse(p,L);
    }
    vector<int> ln(vector<int> p) {
	int t = p.size();
	vector<int> g = Inverse(p) * Derivative(p);
	g = Integral(g);limit(g,t);
	return g;
    }
    vector<int> exp(vector<int> p,int len) {
	vector<int> g,f;
	if(len == 1) {
	    g.pb(1);return g;
	}
	g = exp(p,len >> 1);
	p.resize(len);
	f = g;f.resize(len);
	f = ln(f);
	int L = 1,t = len + g.size() - 1;
	while(L <= t) L <<= 1;
	NTT(p,L,1);NTT(f,L,1);NTT(g,L,1);
	for(int i = 0 ; i < L ; ++i) {
	    g[i] = inc(g[i],mul(g[i],inc(p[i],MOD - f[i])));
	}
	NTT(g,L,-1);
	limit(g,len);
	return g;
    }
    vector<int> Exp(vector<int> p) {
	int L = 1,t = p.size() - 1;
	while(L <= t) L <<= 1;
	return exp(p,L);
    }
    void Print(vector<int> c) {
	for(int i = 0 ; i < c.size() ; ++i) {
	    out(c[i]);space;
	}
	enter;
    }
    void Main() {
	if(Y == 1) {
	    out(fpow(N,2 * N - 4));enter;return;
	}
	int Z = fpow(Y,MOD - 2);Z = inc(Z,MOD - 1);
	fac[0] = 1;
	for(int i = 1 ; i <= N ; ++i) fac[i] = mul(fac[i - 1],i);
	invfac[N] = fpow(fac[N],MOD - 2);
	for(int i = N - 1 ; i >= 0 ; --i) invfac[i] = mul(invfac[i + 1],i + 1);
	inv[1] = 1;
	for(int i = 2 ; i <= N ; ++i) inv[i] = mul(inv[MOD % i],MOD - MOD / i);
	int t = N,l = 1;
	while(l <= t) l <<= 1;
	f.resize(l);
	int invZ = fpow(Z,MOD - 2);
	for(int i = 1 ; i <= N ; ++i) {
	    f[i] = mul(N,N);
	    f[i] = mul(f[i],invZ);
	    f[i] = mul(f[i],fpow(i,i));f[i] = mul(f[i],invfac[i]);
	}
	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]);
	}
	vector<int> a(3),b(3);
	g = Exp(f);
	int ans = g[N];
	ans = mul(ans,fpow(Z,N));
	ans = mul(ans,fpow(fpow(N,MOD - 2),4));
	ans = mul(ans,fac[N]);
	ans = mul(ans,fpow(Y,N));
	out(ans);enter;
    }
}
int main() {
#ifdef ivorysi
    freopen("f1.in","r",stdin);
#endif
    read(N);read(Y);read(op);
    if(op == 0) task0::Main();
    else if(op == 1) task1::Main();
    else if(op == 2) task2::Main();
    return 0;
}