题意
\(n\)个点\(n\)条边的图，有多少种方法给边定向后没有环
Sol
一开始傻了，以为只有一个环。。。实际上N个点N条边还可能是基环树森林。。
做法挺显然的：找出所有的环，设第\(i\)个环的大小为\(w_i\)
\(ans = 2^{N - \sum w_i} \prod (2^{w_i} - 2)\)
最后减掉的2是形成环的情况
#include<bits/stdc++.h>
#define Pair pair<int, int>
#define MP(x, y) make_pair(x, y)
#define fi first
#define se second
#define int long long
#define LL long long
#define Fin(x) {freopen(#x".in","r",stdin);}
#define Fout(x) {freopen(#x".out","w",stdout);}
//#define getchar() (p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 1<<22, stdin), p1 == p2) ? EOF : *p1++)
//char buf[(1 << 22)], *p1 = buf, *p2 = buf;
using namespace std;
const int MAXN = 1e6 + 10, mod = 1e9 + 7, INF = 1e9 + 10;
const double eps = 1e-9, PI = acos(-1);
template <typename A, typename B> inline bool chmin(A &a, B b){if(a > b) {a = b; return 1;} return 0;}
template <typename A, typename B> inline bool chmax(A &a, B b){if(a < b) {a = b; return 1;} return 0;}
template <typename A, typename B> inline LL add(A x, B y) {if(x + y < 0) return x + y + mod; return x + y >= mod ? x + y - mod : x + y;}
template <typename A, typename B> inline void add2(A &x, B y) {if(x + y < 0) x = x + y + mod; else x = (x + y >= mod ? x + y - mod : x + y);}
template <typename A, typename B> inline LL mul(A x, B y) {return 1ll * x * y % mod;}
template <typename A, typename B> inline void mul2(A &x, B y) {x = (1ll * x * y % mod + mod) % mod;}
template <typename A> inline void debug(A a){cout << a << '\n';}
template <typename A> inline LL sqr(A x){return 1ll * x * x;}
inline int read() {
char c = getchar(); int x = 0, f = 1;
while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();}
while(c >= '0' && c <= '9') x = (x * 10 + c - '0') % mod, c = getchar();
return x * f;
}
int N, dep[MAXN], w[MAXN], top, po2[MAXN], vis[MAXN];
vector<int> v[MAXN];
void dfs(int x, int d) {
dep[x] = d; vis[x] = 1;
for(auto &to : v[x]) {
if(!vis[to])dfs(to, d + 1);
else if(vis[to] == 1) w[++top] = dep[x] - dep[to] + 1;
}
vis[x] = 2;
}
signed main() {
N = read();
po2[0] = 1;
for(int i = 1; i <= N; i++) po2[i] = mul(2, po2[i - 1]);
for(int i = 1; i <= N; i++) {
int x = read();
v[i].push_back(x);
}
for(int i = 1; i <= N; i++)
if(!dep[i])
dfs(i, 1);
int sum = 0, res = 1;
for(int i = 1; i <= top; i++) sum += w[i], res = mul(res, po2[w[i]] - 2 + mod);
printf("%d\n", mul(po2[N - sum], res));
return 0;
}