题意
讲不太清楚，看英文吧
cf
做法
在正式开始之前，我们先来玩一玩性质
首先考虑全\(0\)的情况，即本质不同的方案数
性质1：方案数并不为(n-1)!，即方案与结果不为双射
考虑一条边将树分为两个联通块，之间是互不影响的
结论1：方案数为\(\prod\limits (deg_i!)\)
考虑节点\(u\)的父节点与其父亲这条边已经选过
对于每种除\(u\)的儿子外的边固定时，\(u\)的儿子边的全排列决定了其对子树值的影响
ps：实际上这是无根树，但并不影响映射关系
性质2：若\(n\ge 2\)，\(a_i\neq i\)
结论2：若知晓所有限制，\(\sum\limits dis(i,a_i)=2(n-1)\)
具体考虑\(a_v=u\)的限制，有以下结论
- 存在相对顺序边集\((u,t_1)(t_1,t_2)...(t_k,v)\)
- \((u,t_1)\)为\(u\)的第一条边
- \((t_k,v)\)为\(v\)的最后一条边
- \((t_{i-1},t_{i})(t_i,t_{i+1})\)为\(t_i\)严格顺序的边
我们可以依此建立顺序，合法的方案，对于某个点，有以下限制
- 不可破坏简单顺序
- 不可生成环，即为一种非自然的复杂顺序
- 若有严格顺序从第一条边到最后一条边，中间不可漏边
方案数，对于无限制的边随便选择顺序
题外话
这题很妙，虽然容易理解，但需要想满条件还是需要时间的，由于没出现什么算法，强行归到拓扑排序上了
code(std)
#include <bits/stdc++.h>
using namespace std;
const int mod = 1e9 + 7;
const int maxn = 500005;
int n;
vector <int> adj[maxn];
int a[maxn];
int dad[maxn];
int h[maxn];
vector <pair <int, int> > conditions[maxn];
int nxt[maxn], prv[maxn], seen[maxn];
void no(int ncase) {
cerr << ncase << endl;
cout << 0 << endl;
exit(0);
}
void dfs(int u) {
for (auto v: adj[u]) if (v != dad[u]) {
dad[v] = u;
h[v] = h[u] + 1;
dfs(v);
}
}
int cnt; ///total distance, must not be more than 2n-2
void go(int u, int v) {
if (u == v) no(0);
vector <int> from_u, to_v;
///naive LCA works here as long as we exit upon finding a conflict
from_u.push_back(n + 1); ///first edge (fake)
to_v.push_back(n + 2); ///last edge (fake)
while (h[u] > h[v]) {
from_u.push_back(u);
u = dad[u];
}
while (h[v] > h[u]) {
to_v.push_back(v);
v = dad[v];
}
while (u != v) {
from_u.push_back(u);
u = dad[u];
to_v.push_back(v);
v = dad[v];
}
from_u.push_back(u);
from_u.insert(from_u.end(), to_v.rbegin(), to_v.rend());
for (int i = 1; i + 1 < from_u.size(); ++i)
conditions[from_u[i]].push_back({from_u[i-1], from_u[i+1]});
cnt += from_u.size() - 3;
if (cnt > 2 * n - 2) no(0); ///important break
}
int main(void) {
ios_base::sync_with_stdio(0);
cin.tie(NULL);
cin >> n;
for (int i = 1; i < n; ++i) {
int u, v;
cin >> u >> v;
adj[u].push_back(v);
adj[v].push_back(u);
}
for (int i = 1; i <= n; ++i) {
adj[i].push_back(n + 1); ///first edge (fake)
adj[i].push_back(n + 2); ///last edge (fake)
}
for (int i = 1; i <= n; ++i) cin >> a[i];
if (n == 1) {
cout << 1 << endl;
exit(0);
}
dfs(1);
for (int i = 1; i <= n; ++i) if (a[i] != 0) go(i, a[i]);
///answer 0 if:
///1. there are 2 or more incoming/outgoing
///conditions to/from an edge, or
///2. there is a cycle, or
///3. first (n+1) is connected to last (n+2),
///but does not contain all other edges.
int ans = 1;
for (int i = 1; i <= n; ++i) {
///check case 1
for (auto edge: conditions[i]) {
int u = edge.first, v = edge.second;
if (nxt[u] && nxt[u] != v) no(1);
if (prv[v] && prv[v] != u) no(1);
nxt[u] = v;
prv[v] = u;
}
///check case 2
for (auto u: adj[i]) if (!seen[u]) {
seen[u] = 1;
int cur = nxt[u];
while (cur) {
if (cur == u) no(2);
if (seen[cur]) break;
seen[cur] = 1;
cur = nxt[cur];
}
}
///check case 3
if (nxt[n+1]) {
int cur = n + 1, all = 1;
while (cur) {
if (cur == n + 2) break;
++all;
cur = nxt[cur];
}
if (cur == n + 2 && all < adj[i].size()) no(3);
}
///all good - for now
int free = 0;
for (auto u: adj[i]) if (u <= n && !prv[u]) ///fake edges doesn't count
++free;
if (prv[n+2]) --free; ///connected to last edge => not free
for (int j = 1; j <= free; ++j) ans = 1ll * ans * j % mod;
///reset
for (auto u: adj[i]) nxt[u] = prv[u] = seen[u] = 0;
}
///no conflicts
cout << ans << endl;
return 0;
}