题解 心理阴影
确实写出心理阴影了
发现没有祖孙关系的两个点的子树之间互不干扰，于是尝试对这个东西进行 DP
转移的话发现形成的序列中第一个点一定是两个子树的根节点之一
枚举两种情况，发现变成了一个子树和一堆子树，就变成子问题了
先令 \(g_{u, v}\) 表示点 \(v\) 的子树内比 \(u\) 小的点的个数
来康康转移细节
发现如果我定义的 \(f_{u, v}\) 考虑了 \(u, v\) 子树内部的顺序关系的话是没办法转化子问题的（会算重）
（我因为这样定义调了一下午）
于是定义 \(f_{u, v}\) 为在不考虑 \(u, v\) 子树内部顺序的前提下合并两子树的逆序对数平均数
不考虑内部顺序的意思是我不需要知道现在 \(u, v\) 子树内到底是怎么排的
然后考虑转移系数：
考虑合并 \(u, v\) 子树，现在钦定 \(u\) 是拓扑序中的第一个，那么
\[逆序对平均数=\frac{合并 v 和 u 的儿子的逆序对平均数*合并 v 和 u 的儿子的合法方案数}{合并 u 和 v 子树的总方案数}
\]
所以系数就是
\[\begin{aligned}\frac{令u在拓扑序第一个的合法方案数}{合法方案总数}&=\dfrac{\dbinom{siz_u-1+siz_v}{siz_u-1}}{\dbinom{siz_u+siz_v}{siz_u}}\end{aligned}
\]
系数能先化简一下：\(\frac{(siz_u-1+sizv)!}{(siz_u-1)!siz_v!}*\frac{siz_u!siz_v!}{(siz_u+siz_v)!}=\frac{siz_u}{siz_u+siz_v}\)
于是有
枚举兄弟合并，加上每个根节点对其子树的贡献即可
点击查看代码
#include <bits/stdc++.h>
using namespace std;
#define INF 0x3f3f3f3f
#define N 5010
#define ll long long
#define pb push_back
//#define int long long
char buf[1<<21], *p1=buf, *p2=buf;
#define getchar() (p1==p2&&(p2=(p1=buf)+fread(buf, 1, 1<<21, stdin)), p1==p2?EOF:*p1++)
inline int read() {
int ans=0, f=1; char c=getchar();
while (!isdigit(c)) {if (c=='-') f=-f; c=getchar();}
while (isdigit(c)) {ans=(ans<<3)+(ans<<1)+(c^48); c=getchar();}
return ans*f;
}
int n, r;
int head[N], size;
vector<int> to[N];
const ll mod=1e9+7;
struct edge{int to, next;}e[N<<1];
inline void add(int s, int t) {e[++size]={t, head[s]}; head[s]=size;}
inline ll qpow(ll a, ll b) {ll ans=1; for (; b; a=a*a%mod,b>>=1) if (b&1) ans=ans*a%mod; return ans;}
inline ll qinv(ll a) {return qpow(a, mod-2);}
namespace force{
ll ans;
int p[N], pos[N];
bool dfs(int u, int fa) {
for (int i=head[u],v; ~i; i=e[i].next) {
v = e[i].to;
if (v==fa) continue;
if (pos[u]>pos[v]) return 0;
if (!dfs(v, u)) return 0;
}
return 1;
}
void solve() {
for (int i=1; i<=n; ++i) p[i]=i;
int cnt=0;
do {
for (int i=1; i<=n; ++i) pos[p[i]]=i;
if (dfs(r, 0)) {
// cout<<"p: "; for (int i=1; i<=n; ++i) cout<<p[i]<<' '; cout<<endl;
int tem=0;
for (int i=1; i<=n; ++i)
for (int j=i+1; j<=n; ++j)
if (p[i]>p[j]) ++tem;
ans=(ans+tem)%mod; ++cnt;
}
} while (next_permutation(p+1, p+n+1));
printf("%lld\n", ans*qpow(cnt, mod-2)%mod);
}
}
namespace task1{
int bit[N], siz[N], pa[N];
ll fac[N], inv[N], F[N][N], g[N][N], q[N], k[N][N], ans;
inline void upd(int i) {for (; i<=n; i+=i&-i) ++bit[i];}
inline int query(int i) {int ans=0; for (; i; i-=i&-i) ans+=bit[i]; return ans;}
void dfs(int u, int fa) {
for (int i=1; i<=n; ++i) g[i][u]-=query(i-1);
upd(u); siz[u]=1;
for (int i=head[u],v; ~i; i=e[i].next) {
v = e[i].to;
if (v==fa) continue;
pa[v]=u;
dfs(v, u);
siz[u]+=siz[v];
for (int j=1; j<=n; ++j) k[u][j]=(k[u][j]+k[v][j])%mod;
}
for (int i=1; i<=n; ++i) g[i][u]+=query(i-1);
for (int i=1; i<=n; ++i) k[u][i]=(k[u][i]+g[u][i])%mod;
}
void dfs2(int u, int fa) {
for (int i=head[u],v; ~i; i=e[i].next) {
v = e[i].to;
if (v==fa) continue;
dfs2(v, u);
for (int j=1; j<=n; ++j) k[u][j]=(k[u][j]+k[v][j])%mod;
}
for (int i=1; i<=n; ++i) k[u][i]=(k[u][i]+g[u][i])%mod;
}
ll f(int , int ) ;
ll calc(int r, int fa) {
ll ans=g[r][r];
for (int i=0; i<to[r].size(); ++i)
for (int j=0; j<to[r].size(); ++j) if (j!=i) {
int u=to[r][i], v=to[r][j];
if (u!=fa&&v!=fa) ans=(ans+k[u][v])%mod;
}
return ans;
}
ll f(int u, int v) {
cout<<"f: "<<u<<' '<<v<<endl;
if (u>v) swap(u, v);
if (~F[u][v]) return F[u][v];
ll* t=&F[u][v]; *t=0;
ll t1=g[u][v], t2=g[v][u];
for (int i=head[u],w; ~i; i=e[i].next) {
w = e[i].to;
if (w!=pa[u]) t1=(t1+f(w, v))%mod;
}
for (int i=head[v],w; ~i; i=e[i].next) {
w = e[i].to;
if (w!=pa[v]) t2=(t2+f(w, u))%mod;
}
if (u==3&&v==5) cout<<t1<<' '<<t2<<endl;
t1=t1*siz[u]%mod; t2=t2*siz[v]%mod;
if (u==3&&v==5) cout<<t1<<' '<<t2<<endl;
cout<<"f: "<<u<<' '<<v<<' '<<(t1+t2)*qinv(siz[u]+siz[v])%mod<<endl;
// ll tem=fac[siz[u]+siz[v]]*inv[siz[u]]%mod*inv[siz[v]]%mod;
return *t=((t1+t2)*qinv(siz[u]+siz[v])+(calc(u, pa[u])+calc(v, pa[v])))%mod;
}
void solve() {
dfs(r, 0); dfs2(r, 0);
memset(F, -1, sizeof(F));
fac[0]=fac[1]=1; inv[0]=inv[1]=1;
for (int i=2; i<=n; ++i) fac[i]=fac[i-1]*i%mod;
for (int i=2; i<=n; ++i) inv[i]=(mod-mod/i)*inv[mod%i]%mod;
for (int i=2; i<=n; ++i) inv[i]=inv[i-1]*inv[i]%mod;
// cout<<"g"<<endl; for (int i=1; i<=n; ++i) {for (int j=1; j<=n; ++j) cout<<g[i][j]<<' '; cout<<endl;}
while (to[r].size()-(pa[r]>0)==1) {
ans=(ans+g[r][r]);
r=to[r][0]==pa[r]?to[r][1]:to[r][0];
}
// cout<<"r: "<<r<<endl;
// for (int i=0; i<to[r].size(); ++i)
// for (int j=i+1; j<to[r].size(); ++j) {
// int u=to[r][i], v=to[r][j];
// if (u==pa[r]||v==pa[r]) continue;
// ans=(ans+f(u, v))%mod;
// }
printf("%lld\n", (ans+calc(r, pa[r]))%mod);
// cout<<f(2, 5)<<endl;
// cout<<calc(5, pa[5])<<endl;
// cout<<calc(2, pa[2])<<endl;
cout<<f(3, 5)<<endl;
cout<<f(2, 5)<<endl;
// cout<<k[5][3]<<endl;
cout<<calc(3, pa[3])<<endl;
cout<<9*qinv(4)%mod<<endl;
cout<<3*qinv(2)%mod<<endl;
}
}
namespace task{
int bit[N], siz[N], pa[N];
ll fac[N], inv[N], F[N][N], g[N][N], q[N], k[N][N], ans;
inline void upd(int i) {for (; i<=n; i+=i&-i) ++bit[i];}
inline int query(int i) {int ans=0; for (; i; i-=i&-i) ans+=bit[i]; return ans;}
void dfs(int u, int fa) {
siz[u]=1;
for (int i=head[u],v; ~i; i=e[i].next) {
v = e[i].to;
if (v==fa) continue;
pa[v]=u;
dfs(v, u);
siz[u]+=siz[v];
for (int j=1; j<=n; ++j) g[u][j]+=g[v][j];
}
for (int i=1; i<=n; ++i) g[u][i]+=(i>u);
}
ll f(int u, int v) {
if (u>v) swap(u, v);
if (~F[u][v]) return F[u][v];
ll* t=&F[u][v]; *t=0;
ll t1=g[v][u], t2=g[u][v];
for (int i=head[u],w; ~i; i=e[i].next) {
w = e[i].to;
if (w!=pa[u]) t1=(t1+f(w, v))%mod;
}
for (int i=head[v],w; ~i; i=e[i].next) {
w = e[i].to;
if (w!=pa[v]) t2=(t2+f(w, u))%mod;
}
t1=t1*siz[u]%mod; t2=t2*siz[v]%mod;
return *t=(t1+t2)*inv[siz[u]+siz[v]]%mod;
}
void solve() {
dfs(r, 0);
memset(F, -1, sizeof(F));
fac[0]=fac[1]=1; inv[0]=inv[1]=1;
for (int i=2; i<=n; ++i) fac[i]=fac[i-1]*i%mod;
for (int i=2; i<=n; ++i) inv[i]=(mod-mod/i)*inv[mod%i]%mod;
// cout<<"g"<<endl; for (int i=1; i<=n; ++i) {for (int j=1; j<=n; ++j) cout<<g[i][j]<<' '; cout<<endl;}
for (int i=1; i<=n; ++i)
for (int j=i+1; j<=n; ++j)
if (pa[i]==pa[j])
ans=(ans+f(i, j))%mod;
for (int i=1; i<=n; ++i) ans=(ans+g[i][i])%mod;
// cout<<f(2, 5)<<endl;
printf("%lld\n", ans);
}
}
signed main()
{
freopen("nightmare.in", "r", stdin);
freopen("nightmare.out", "w", stdout);
n=read(); r=read();
memset(head, -1, sizeof(head));
for (int i=1,u,v; i<n; ++i) {
u=read(); v=read();
add(u, v); add(v, u);
to[u].pb(v); to[v].pb(u);
}
// force::solve();
task::solve();
return 0;
}
