Problem G. Depth-First Search
PS:官方题解挺详细的,就是自己写有点难度,在DFS以a[1]为根的树中,要向下传递上一层的合法方案数。也就是 v = v * sum * fac[i](v就是上一层的合法方案数),吐槽一下,计数类的问题是真的麻烦,不重不漏。
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
#include<bits/stdc++.h> #define ll long long #define P pair<int, int> #define pb push_back #define mp make_pair #define pp pop_back #define lson root << 1 #define INF (int)2e9 + 7 #define rson root << 1 | 1 #define LINF (unsigned long long int)1e18 #define sc(x) scanf("%d", &x) #define pr(x) printf("%d\n", x) #define mem(arry, in) memset(arry, in, sizeof(arry)) #define PI acos(0.5) * 3 #define EPS 0.00000001 using namespace std; inline void upd(int&x, int y) { x < y && (x = y); } const int N = 1000005; const int mod = 1000000007; vector<int> G[N]; int powi(int a, int b) { int c = 1; for (; b; a = 1ll * a * a % mod, b >>= 1) if (b & 1) c = 1ll * c * a % mod; return c; } inline int mul(int a, int b, int c) { return 1ll * a * b % mod * c % mod; } inline void init(int n) { for (int i = 1; i <= n; ++i) G[i].clear(); } struct Tree { int n; vector<int> T; void init(int _n) { n = _n; T.resize(n + 1); for (int i = 0; i <= n; ++i) T[i] = 0; } void add(int pos, int x) { for (int i = pos; i <= n; i += i & -i) T[i] += x; } int sum(int pos) { if (pos > n) pos = n; int res = 0; for (int i = pos; i; i -= i & -i) res += T[i]; return res; } }bit[N]; int T, n, ans, id, d; int a[N], fac[N], inv[N], f[N], invf[N]; void Inite() { int mx = 1000000; fac[0] = 1; for (int i = 1; i <= mx; ++i) fac[i] = 1ll * fac[i - 1] * i % mod; inv[mx] = powi(fac[mx], mod - 2); for (int i = mx - 1; i; --i) inv[i] = 1ll * inv[i + 1] * (i + 1) % mod; } int solve(int u, int v) { id++; int sum = 1; for (auto tp : G[u]) sum = 1ll * sum * f[tp] % mod; for (int i = G[u].size() - 1; ~i; --i) { int nxt = lower_bound(G[u].begin(), G[u].end(), a[id + 1]) - G[u].begin(); int cnt = bit[u].sum(nxt); ans = (0ll + ans + 1ll * mul(sum, fac[i], v) * cnt % mod) % mod; if (nxt == G[u].size() || G[u][nxt] != a[id + 1]) return 1; bit[u].add(nxt + 1, -1); sum = 1ll * sum * invf[G[u][nxt]] % mod; if (solve(G[u][nxt], mul(v, fac[i], sum))) return 1; } return 0; } void DFS(int u, int p) { if (p > 0) G[u].erase(find(G[u].begin(), G[u].end(), p)); f[u] = fac[G[u].size()]; bit[u].init(G[u].size()); for (int i = 0; i < G[u].size(); ++i) bit[u].add(i + 1, 1); if(!G[u].empty()) { sort(G[u].begin(), G[u].end()); for (auto v : G[u]) { DFS(v, u); f[u] = 1ll * f[u] * f[v] % mod; } } invf[u] = powi(f[u], mod - 2); } int main() { Inite(); sc(T); while(T--) { sc(n); init(n); for (int i = 0; i < n; ++i) sc(a[i]); for (int i = 1; i < n; ++i) { int u, v; sc(u), sc(v); G[u].pb(v); G[v].pb(u); } a[n] = 0, d = 1, ans = 0, id = -1; for (int i = 1; i <= n; ++i) d = 1ll * d * fac[G[i].size() - 1] % mod; for (int i = 1; i < a[0]; ++i) ans = (0ll + ans + 1ll * d * G[i].size() % mod) % mod; DFS(a[0], -1); solve(a[0], 1); printf("%d\n", ans); } return 0; }
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