题目链接：https://vjudge.net/problem/Gym-101234D
题目大意：给一颗树，共N个点，每次随机选择一个点，得分加上该点所在树的大小，然后删除这个点，断开与其相连的所有边，问删完所有点所获得的 期望得分乘N！ 是多少。
emmm被这个乘N! 给坑了，以为这是个假的期望题目，因为全排列共N!种， 期望乘N! 其实就是全排列的和，觉得这是出题人留的解题思路.....然后就去考虑和的问题了，从此走上了不归路。
看了题解后发现，这个N! 一点用都没有啊啊啊啊啊啊啊。
重新用期望的方式开始思考，我们考虑节点u对节点v的贡献，u对v有贡献当且仅当u是u->v这条链上第一个删除的点，概率为1/len(len是该链上点的个数，包含端点）。
然后就变成了求树上每种长度（长度定义为路径上点数）的路径条数的模板，除了长度为1的以外，都需要算2次。
点分治的过程中使用fft合并就可以了。
复杂度O (N * logN * logN)
#include <bits/stdc++.h>
#define ll long long
#define rep(i, a, b) for(int i = (a); i <= (b); i++)
#define pb push_back
#define ll long long
using namespace std;
const int M = 1e5+1000;
const int mod = 1e9+7;
const double PI = acos(-1.0);
struct Complex {
double x, y;
Complex(double _x=0.0,double _y = 0.0) {
x = _x;
y = _y;
}
Complex operator -(const Complex &b) const {
return Complex(x-b.x, y-b.y);
}
Complex operator +(const Complex &b) const {
return Complex(x+b.x, y+b.y);
}
Complex operator *(const Complex &b) const {
return Complex(x*b.x-y*b.y, x*b.y+y*b.x);
}
};
void change(Complex y[], int len) {
int i,j,k;
for(i = 1, j = len/2; i < len-1; i++) {
if(i < j) swap(y[i],y[j]);
k = len/2;
while(j >= k) {
j -= k;
k /= 2;
}
if(j < k) j += k;
}
}
void fft(Complex y[], int len, int on) {
change(y, len);
for(int h = 2; h <= len; h <<= 1) {
Complex wn(cos(-on*2*PI/h),sin(-on*2*PI/h));
for(int j = 0; j <= len; j+= h) {
Complex w(1,0);
for(int k = j; k < j+h/2; k++) {
Complex u = y[k];
Complex t = w*y[k+h/2];
y[k] = u+t;
y[k+h/2] = u-t;
w = w*wn;
}
}
}
if(on == -1)
for(int i = 0; i < len; i++)
y[i].x /= len;
}
const int N = 800010; //表达式长度的八倍
Complex x1[N],x2[N];
ll sum[N],Tnum[N],num[N];
void mul(ll a[],ll b[],ll sum[],int n,int m) {
int len1 = n+1;
int len2 = m+1;
int len = 1;
while(len < len1*2 || len < len2*2) len <<= 1;
for(int i = 0; i < len1; i++) x1[i] = Complex(a[i],0);
for(int i = len1; i < len; i++) x1[i] = Complex(0,0);
for(int i = 0; i < len2; i++) x2[i] = Complex(b[i],0);
for(int i = len2; i < len; i++) x2[i] = Complex(0,0);
fft(x1,len,1);
fft(x2,len,1);
for(int i = 0; i < len; i++) x1[i] = x1[i]*x2[i];
fft(x1,len,-1);
for(int i = 0; i < len1+len2-1; i++) sum[i] += (ll)(x1[i].x+0.5);
}
//--------------------------------------------------
int n;
struct node {
int v,nxt;
}edge[2*M];
int tot,head[M];
void ae(int u,int v) {
edge[++tot] = node{v,head[u]};
head[u] = tot;
}
void init(int n) {
tot = 0;
rep(i, 1, n) head[i] = -1;
}
int siz[M],Root,wt[M],Tsiz,len1,len2;
bool vis[M];
void GetRoot(int u,int f) {
siz[u] = 1;
wt[u] = 0;
for(int i = head[u]; ~i ; i = edge[i].nxt) {
int v = edge[i].v;
if(v==f||vis[v]) continue;
GetRoot(v,u);
siz[u] += siz[v];
wt[u] = max(wt[u],siz[v]);
}
wt[u] = max(wt[u],Tsiz-siz[u]);
if(wt[Root]>wt[u]) Root = u;
}
void dfs(int u,int f,int dis) {
num[dis]++;
len2 = max(len2,dis);
for(int i = head[u]; ~i ; i = edge[i].nxt) {
int v = edge[i].v;
if(v==f||vis[v]) continue;
dfs(v,u,dis+1);
}
}
void calc(int u) {
len1 = 1; Tnum[1] = 1;
for(int i = head[u]; ~i ; i = edge[i].nxt) {
int v = edge[i].v;
if(vis[v]) continue;
len2 = 1;
dfs(v,u,1);
mul(Tnum,num,sum,len1,len2);
rep(i, 1 ,len2) {
Tnum[i+1] += num[i];
num[i] = 0;
}
len1 = max(len1,len2+1);
}
rep(i, 0, len1) Tnum[i] = 0;
}
void divide(int u) {
calc(u);
vis[u] = 1; //删掉该点
for(int i = head[u]; ~i ; i = edge[i].nxt) {
int v = edge[i].v;
if(vis[v]) continue;
Root = 0,Tsiz = siz[v];
GetRoot(v,0);
divide(Root);
}
}
ll pow_mod(ll n,ll m) {
ll ans = 1;
while(m) {
if(m&1) ans = ans*n%mod;
n = n*n%mod;
m >>= 1;
}
return ans;
}
int main() {
//freopen("a.txt","r",stdin);
//ios::sync_with_stdio(0);
scanf("%d",&n);
init(n);
rep(i, 1, n-1) {
int u,v;
scanf("%d%d",&u,&v);
ae(u,v);
ae(v,u);
}
rep(i, 1, n) vis[i] = 0;
wt[0] = 1e9,Tsiz = n,GetRoot(1,0);
divide(Root);
ll ans = 0;
sum[1] = n;
rep(i, 2, n) sum[i] = sum[i]*2%mod;
rep(i, 1, n) ans = (ans + sum[i] * pow_mod(i,mod-2))%mod;
rep(i, 1, n) ans = ans*i%mod;
printf("%lld",ans);
return 0;
}