BZOJ 4011: [HNOI2015]落忆枫音

如果不新加边，那么给每个点选一条入边即可，答案为 \(\prod \limits_{i=2}^n in\_degree_i\)
加入新边后，仍按上面的方法统计答案，但是原图中可能出现了环，如果出现环，那么边 \(x \to y\) 一定就在环上
那么就统计选了 \(y \to x\) 的路径的方案数
设 \(dp_i\) 表示选了 \(y \to i\) 的方案数
\(dp_i = \frac{\sum \limits_{(j, i) \in E} dp_j}{in\_degree_i}\)
\(dp_y = \frac{\prod \limits_{i=2}^n in\_degree_i}{in\_degree_y}\)

#include <bits/stdc++.h>
#define pb push_back
#define fi first
#define se second
#define pii pair<int, int>
#define pli pair<ll, int>
#define lp p << 1
#define rp p << 1 | 1
#define mid ((l + r) / 2)
#define lowbit(i) ((i) & (-i))
typedef long long ll;
typedef unsigned long long ull;
typedef double db;
#define rep(i,a,b) for(int i=a;i<b;i++)
#define per(i,a,b) for(int i=b-1;i>=a;i--)
#define Edg int ccnt=1,head[N],to[E],ne[E];void addd(int u,int v){to[++ccnt]=v;ne[ccnt]=head[u];head[u]=ccnt;}void add(int u,int v){addd(u,v);addd(v,u);}
#define Edgc int ccnt=1,head[N],to[E],ne[E],c[E];void addd(int u,int v,int w){to[++ccnt]=v;ne[ccnt]=head[u];c[ccnt]=w;head[u]=ccnt;}void add(int u,int v,int w){addd(u,v,w);addd(v,u,w);}
#define es(u,i,v) for(int i=head[u],v=to[i];i;i=ne[i],v=to[i])
const int MOD = 1e9 + 7;
void M(int &x) {if (x >= MOD)x -= MOD; if (x < 0)x += MOD;}
int qp(int a, int b = MOD - 2) {int ans = 1; for (; b; a = 1LL * a * a % MOD, b >>= 1)if (b & 1)ans = 1LL * ans * a % MOD; return ans % MOD;}
template<class T>T gcd(T a, T b) { while (b) { a %= b; std::swap(a, b); } return a; }
template<class T>bool chkmin(T &a, T b) { return a > b ? a = b, 1 : 0; }
template<class T>bool chkmax(T &a, T b) { return a < b ? a = b, 1 : 0; }
char buf[1 << 21], *p1 = buf, *p2 = buf;
inline char getc() {
	return p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 1 << 21, stdin), p1 == p2) ? EOF : *p1++;
}
inline int _() {
	int x = 0, f = 1; char ch = getc();
	while (ch < '0' || ch > '9') { if (ch == '-') f = -1; ch = getc(); }
	while (ch >= '0' && ch <= '9') { x = x * 10ll + ch - 48; ch = getc(); }
	return x * f;
}

const int N = 1e5 + 7, E = 2e5 + 7;
int n, m, x, y, in[N], inn[N], inv[E];
int que[N], l, r, dp[N];
Edg

int solve() {
	inv[1] = 1;
	rep (i, 2, m + 5) inv[i] = 1LL * (MOD - MOD / i) * inv[MOD % i] % MOD;
	int l = 0, r = 0;
	rep (i, 1, n + 1) if (!in[i]) que[r++] = i;
	while (l ^ r) {
		int u = que[l++];
		dp[u] = 1LL * dp[u] * inv[inn[u]] % MOD;
		es (u, i, v) {
			M(dp[v] += dp[u]);
			if (!--in[v]) que[r++] = v;
		}
	}
	return dp[x];
}

int main() {
#ifdef LOCAL
	freopen("ans.out", "w", stdout);
#endif
	n = _(), m = _(), x = _(), y = _();
	inn[y]++;
	rep (i, 0, m) {
		int u = _(), v = _();
		addd(u, v);
		in[v]++;
		inn[v]++;
	}
	int ans = 1;
	rep (i, 2, n + 1) ans = 1LL * ans * inn[i] % MOD;
	if (y != 1) dp[y] = ans, M(ans -= solve());
	printf("%d\n", ans);
#ifdef LOCAL
	printf("%.10f\n", (db)clock() / CLOCKS_PER_SEC);
#endif
	return 0;
}