题面

洛谷

题解

20pts

枚举每一条边是否在树中即可。

另10pts

我们考虑一张\(DAG\)中构成树的方法数，每个点选一个父亲即可，那么有

\(deg_i\)表示点\(i\)的入度，其中\(deg_1=1\)。

\(100pts\)

考虑在上面的基础上容斥，

考虑连\(y\rightarrow x\)后出现一个环的情况数，其实就是环上的点固定了父亲，

那么最后答案就是

代码

#include <iostream>

#include <cstdio>

#include <cstdlib>

#include <cstring>

#include <cmath>

#include <algorithm>

#include <vector>

#include <queue>

using namespace std;

inline int gi() {

    register int data = 0, w = 1;

    register char ch = 0;

    while (!isdigit(ch) && ch != '-') ch = getchar();

    if (ch == '-') w = -1, ch = getchar();

    while (isdigit(ch)) data = 10 * data + ch - '0', ch = getchar();

    return w * data;

}

const int Mod = 1e9 + 7;

const int MAX_N = 1e5 + 5, MAX_M = 2e5 + 5;

int fpow(int x, int y) {

	int res = 1;

	while (y) {

		if (y & 1) res = 1ll * res * x % Mod;

		x = 1ll * x * x % Mod;

		y >>= 1;

	}

	return res;

}

struct Graph { int to, next; } e[MAX_M]; int fir[MAX_N], e_cnt;

void clearGraph() { memset(fir, -1, sizeof(fir)); e_cnt = 0; }

void Add_Edge(int u, int v) { e[e_cnt] = (Graph){v, fir[u]}, fir[u] = e_cnt++; }

int N, M, sx, sy, deg[MAX_N], f[MAX_N];

bool vis[MAX_N];

void dfs(int x) {

	if (vis[x]) return ; vis[x] = 1;

	if (sx == x) return (void)(f[x] = 1ll * fpow(deg[x], Mod - 2));

	for (int i = fir[x]; ~i; i = e[i].next) dfs(e[i].to), f[x] = (f[x] + f[e[i].to]) % Mod;

	f[x] = 1ll * f[x] * fpow(deg[x], Mod - 2) % Mod;

}

int main () {

#ifndef ONLINE_JUDGE

    freopen("cpp.in", "r", stdin);

#endif

	clearGraph();

	N = gi(), M = gi(), sx = gi(), sy = gi();

	for (int u, v, i = 1; i <= M; i++) u = gi(), v = gi(), Add_Edge(u, v), ++deg[v];

	++deg[1];

	int ans = 1, res = 1;

	for (int i = 1; i <= N; i++) {

		if (i == sy) ans = 1ll * ans * (deg[i] + 1) % Mod;

		else ans = 1ll * ans * deg[i] % Mod;

		res = 1ll * res * deg[i] % Mod;

	}

	dfs(sy); ans = (ans - 1ll * res * f[sy] % Mod + Mod) % Mod;

	printf("%d\n", ans);

    return 0;

}