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
\[Ans=\prod_{i=1}^{n} deg_i\]

\(deg_i\)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

\[ \prod_{i=1}^{n} deg_i-\frac {\prod_{i=1}^{n} deg_i}{\prod_{j\in loop} deg_j} \]

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);
#endifclearGraph(); 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;
}