Zhang_RQ题解（本篇仅概述）
前言
有进步，只做了半天。。。。
一道具有极强综合性的数数好题！
强大的多合一题目
精确地数学推导和耐心。
有套路又不失心意。
融合了：
算法：
prufer序列及其扩展
树形Dp
容斥或者二项式定理
EGF
多项式Exp
先要会：
省选模拟赛第四轮 B——O(n^4)->O(n^3)->O(n^2)
然后开始刚题。
就是：
求各种情况下的y^(n-|T1∩T2|)的和
OP=0
哈希。
OP=1
枚举交集，统计符合的T2树的个数
prufer序列森林版扩展
但是会算重，连回去会多
考虑重组贡献
每个实际的$y^k$会被$k$小的时候计算，$y^k=(y+1-1)^k=∑C(j,i)(y-1)^i$
发现，只要把贡献的$y$成$y-1$，就大功告成啦！
$\Pi szi$可以套路地组合意义成为每个连通块选择一个的方案数，然后就$f[i][0/1]$就O(n)辣
注意每次选择交的边还有贡献
这里，第二个$y^{-1}$运算的时候，要变成$(y^{-1}-1)$
OP=2
还是枚举交集的边
发现和交集大小密切相关，所以枚举交集大小$s$
然后$gs$设出来，
要枚举边再考虑连通块，不妨枚举连通块考虑贡献的边集
连通块的分配$\frac{n!}{(n-s)!}$,连通块内的连边$a^{a-2}$，连通块之间连边$\Pi a_i *n^{m-2}$
一些东西要平方
把$gs$带回去，然后无关的都提到前面去
把枚举连通块，变成EGF
然后再变成多项式指数函数的形式，大功告成！
#include<bits/stdc++.h> #define reg register int #define il inline #define fi first #define se second #define mk(a,b) make_pair(a,b) #define numb (ch^'0') using namespace std; typedef long long ll; template<class T>il void rd(T &x){ char ch;x=0;bool fl=false; while(!isdigit(ch=getchar()))(ch=='-')&&(fl=true); for(x=numb;isdigit(ch=getchar());x=x*10+numb); (fl==true)&&(x=-x); } template<class T>il void output(T x){if(x/10)output(x/10);putchar(x%10+'0');} template<class T>il void ot(T x){if(x<0) putchar('-'),x=-x;output(x);putchar(' ');} template<class T>il void prt(T a[],int st,int nd){for(reg i=st;i<=nd;++i) ot(a[i]);putchar('\n');} namespace Miracle{ const int N=1e5+5; const int mod=998244353; const int G=3; const int GI=332748118; int n,Y; int mul(int x,int y){return (ll)x*y%mod;} int ad(int x,int y){return x+y>=mod?x+y-mod:x+y;} il int sub(int x,int y){return ad(x,mod-y);} int qm(int x,int y){ int ret=1; while(y){ if(y&1) ret=mul(ret,x); x=mul(x,x); y>>=1; } return ret; } namespace sol0{ map<pair<int,int>,int>mp; int cnt; void main(){ int x,y; for(reg i=1;i<n;++i){ rd(x);rd(y);if(x>y) swap(x,y);mp[mk(x,y)]=1; } for(reg i=1;i<n;++i){ rd(x);rd(y);if(x>y) swap(x,y);cnt+=mp[mk(x,y)]; } printf("%d\n",qm(Y,n-cnt)); } } namespace sol1{ ll f[N][2]; int z; int cos,ivc; struct node{ int nxt,to; }e[2*N]; int hd[N],cnt; void add(int x,int y){ e[++cnt].nxt=hd[x]; e[cnt].to=y; hd[x]=cnt; } void dfs(int x,int fa){ f[x][0]=f[x][1]=1; for(reg i=hd[x];i;i=e[i].nxt){ int y=e[i].to; if(y==fa) continue; dfs(y,x); //jiao and no ll od0=f[x][0],od1=f[x][1]; f[x][0]=ad(mul(od0,f[y][1]),mul(od0,mul(f[y][0],cos))); f[x][1]=ad(mul(od1,f[y][1]),ad(mul(od0,mul(f[y][1],cos)),mul(od1,mul(f[y][0],cos)))); } } void main(){ int x,y; for(reg i=1;i<n;++i){ rd(x);rd(y);add(x,y);add(y,x); } if(Y==1){ printf("%d\n",qm(n,n-2)); return; } z=ad(qm(Y,mod-2),mod-1); cos=mul(z,qm(n,mod-2)); ivc=qm(cos,mod-2); // cout<<" cos "<<cos<<" ivc "<<ivc<<endl; dfs(1,0); ll ans=0; ans=mul(mul(qm(n,n-2),qm(Y,n)),f[1][1]); cout<<ans; } } namespace sol2{ struct Poly{ vector<int>f; Poly(){f.clear();} il int &operator[](const int &x){return f[x];} il const int &operator[](const int &x) const {return f[x];} il void resize(const int &n){f.resize(n);} il int size() const {return f.size();} il void cpy(Poly &b){f.resize(b.size());for(reg i=0;i<(int)f.size();++i)f[i]=b[i];} il void rev(){reverse(f.begin(),f.end());} il void clear(){f.clear();} il void read(const int &n){f.resize(n);for(reg i=0;i<n;++i)rd(f[i]);} il void out() const {for(reg i=0;i<(int)f.size();++i)ot(f[i]);putchar('\n');} }R; il int init(const int &n){int m;for(m=1;m<n;m<<=1);return m;} il void rev(Poly &f){ int lp=f.size(); if(R.size()!=f.size()) { R.resize(f.size()); for(reg i=0;i<lp;++i){ R[i]=(R[i>>1]>>1)|((i&1)?lp>>1:0); } } for(reg i=0;i<lp;++i){ if(i<R[i]) swap(f[i],f[R[i]]); } } il void NTT(Poly &f,int c){ int n=f.size();rev(f); for(reg p=2;p<=n;p<<=1){ int gen=(c==1)?qm(G,(mod-1)/p):qm(GI,(mod-1)/p); for(reg l=0;l<n;l+=p){ int buf=1; for(reg k=l;k<l+p/2;++k){ int tmp=mul(f[k+p/2],buf); f[k+p/2]=sub(f[k],tmp); f[k]=ad(f[k],tmp); buf=mul(buf,gen); } } } if(c==-1){ int iv=qm(n,mod-2);for(reg i=0;i<n;++i) f[i]=mul(f[i],iv); } } il Poly Inv(const Poly &f,int n){ if(n==1){ Poly g;g.resize(1);g[0]=qm(f[0],mod-2);return g; } Poly g=Inv(f,(n+1)>>1),t; int m=init(n*2); t.resize(m); for(reg i=0;i<n;++i)t[i]=f[i]; g.resize(m); NTT(g,1);NTT(t,1); for(reg i=0;i<m;++i)g[i]=mul(sub(2,mul(g[i],t[i])),g[i]); NTT(g,-1);g.resize(n); return g; } il void operator *=(Poly &f,Poly g){ int st=f.size()+g.size()-1; int len=init(f.size()+g.size()-1);f.resize(len);g.resize(len); NTT(f,1);NTT(g,1);for(reg i=0;i<len;++i) f[i]=mul(f[i],g[i]); NTT(f,-1); f.resize(st); } il void operator *=(Poly &f,const int &c){for(reg i=0;i<f.size();++i) f[i]=mul(f[i],c);} il Poly operator *(Poly f,const Poly &g){f*=g;return f;} il Poly operator *(Poly f,const int &c){for(reg i=0;i<f.size();++i) f[i]=mul(f[i],c);return f;} il void operator +=(Poly &f,const Poly &g){for(reg i=0;i<f.size();++i) f[i]=ad(f[i],g[i]);} il void operator +=(Poly &f,const int &c){f[0]=ad(f[0],c);} il Poly operator +(Poly f,const Poly &g){for(reg i=0;i<f.size();++i) f[i]=ad(f[i],g[i]);return f;} il Poly operator +(Poly f,const int &c){f[0]=ad(f[0],c);return f;} il void operator -=(Poly &f,const Poly &g){for(reg i=0;i<f.size();++i) f[i]=sub(f[i],g[i]);} il void operator -=(Poly &f,const int &c){f[0]=sub(f[0],c);} il Poly operator -(Poly f,const Poly &g){for(reg i=0;i<f.size();++i) f[i]=sub(f[i],g[i]);return f;} il Poly operator -(Poly f,const int &c){f[0]=sub(f[0],c);return f;} il Poly operator ~(const Poly &f){return Inv(f,f.size());} il Poly operator /(Poly f,Poly g){int len=f.size()-g.size()+1;f.rev();g.rev();g.resize(len);f=f*(~g);f.resize(len);f.rev();return f;} il Poly operator %(Poly f,Poly g){Poly s=f/g;f=f-g*s;f.resize(g.size()-1);return f;} il Poly Inter(Poly f){int st=f.size();f.resize(st+1);for(reg i=st;i>=1;--i){f[i]=mul(f[i-1],qm(i,mod-2));}f[0]=0;return f;} il Poly Diff(Poly f){int st=f.size();for(reg i=0;i<st-1;++i) f[i]=mul(f[i+1],(i+1));f.resize(st-1);return f;} il Poly Ln(const Poly &f){Poly g=Diff(f),h=(~f);g=g*h;g.resize(f.size()-1);return Inter(g);} il Poly Exp(const Poly &f,int n){ if(n==1){ Poly g;g.resize(1);g[0]=1; return g; } Poly g=Exp(f,(n+1)>>1); g.resize(n); g=g*(((Ln(g)*(mod-1))+1)+f); g.resize(n); return g; } il Poly Exp(const Poly &f){ return Exp(f,f.size()); } int jie[N],inv[N]; void main(){ if(Y==1){ printf("%d\n",mul(qm(n,n-2),qm(n,n-2))); return; } int z=ad(qm(Y,mod-2),mod-1); Poly g; g.resize(n+1); int ivz=qm(z,mod-2); jie[0]=1; for(reg i=1;i<=n;++i) jie[i]=mul(jie[i-1],i); inv[n]=qm(jie[n],mod-2); for(reg i=n-1;i>=0;--i) inv[i]=mul(inv[i+1],i+1); for(reg j=1;j<=n;++j){ g[j]=mul(mul(mul(mul(n,n),ivz),inv[j]),qm(j,j)); } g=Exp(g); ll ans=mul(mul(mul(mul(qm(Y,n),jie[n]),qm(qm(n,mod-2),4)),qm(z,n)),g[n]); cout<<ans; } } int main(){ int op; rd(n);rd(Y);rd(op); if(op==0) sol0::main(); else if(op==1) sol1::main(); else sol2::main(); return 0; } } signed main(){ Miracle::main(); return 0; } /* Author: *Miracle* Date: 2019/4/12 15:09:42 */