Codeforces Round #369 (Div. 2) D. Directed Roads 数学

D. Directed Roads

题目连接：

http://www.codeforces.com/contest/711/problem/D

Description

ZS the Coder and Chris the Baboon has explored Udayland for quite some time. They realize that it consists of n towns numbered from 1 to n.

There are n directed roads in the Udayland. i-th of them goes from town i to some other town ai (ai ≠ i). ZS the Coder can flip the direction of any road in Udayland, i.e. if it goes from town A to town B before the flip, it will go from town B to town A after.

ZS the Coder considers the roads in the Udayland confusing, if there is a sequence of distinct towns A1, A2, ..., Ak (k > 1) such that for every 1 ≤ i < k there is a road from town Ai to town Ai + 1 and another road from town Ak to town A1. In other words, the roads are confusing if some of them form a directed cycle of some towns.

Now ZS the Coder wonders how many sets of roads (there are 2n variants) in initial configuration can he choose to flip such that after flipping each road in the set exactly once, the resulting network will not be confusing.

Note that it is allowed that after the flipping there are more than one directed road from some town and possibly some towns with no roads leading out of it, or multiple roads between any pair of cities.

Input

The first line of the input contains single integer n (2 ≤ n ≤ 2·105) — the number of towns in Udayland.

The next line contains n integers a1, a2, ..., an (1 ≤ ai ≤ n, ai ≠ i), ai denotes a road going from town i to town ai.

Output

Print a single integer — the number of ways to flip some set of the roads so that the resulting whole set of all roads is not confusing. Since this number may be too large, print the answer modulo 109 + 7.

Sample Input

3
2 3 1

Sample Output

6

Hint

题意

给你一个\(n\)点\(n\)边的无向图，你可以翻转任意几条边，但是每条边只能翻转一次，问你有多少种方案，使得这个图不存在环。

题解：

由于n点n边，所以不存在环套环的情况。

对于每一个环，我们都有2^{n-2种方案使得这个环不存在，就是2}n减去全部翻转，和全部不翻转的方案。

然后不在环上的边，爱翻翻，乘上去这个答案就好了。

代码

#include<bits/stdc++.h>
using namespace std;
const int mod = 1e9+7;
const int maxn = 2e5+7;
long long f[maxn];
int a[maxn],n,cnt;
int vis[maxn];
int main()
{
    scanf("%d",&n);
    for(int i=1;i<=n;i++)
        scanf("%d",&a[i]);
    f[0]=1;
    for(int i=1;i<=n;i++)
        f[i]=f[i-1]*2ll%mod;
    int num = n;
    long long ans = 1;
    for(int i=1;i<=n;i++)
    {
        if(!vis[i])
        {
            int x=i;
            int now=cnt;
            while(vis[x]==0)
            {
                vis[x]=++cnt;
                x=a[x];
            }
            if(vis[x]>now)
            {
                int pnum=cnt-vis[x]+1;
                num=num-pnum;
                ans=ans*(f[pnum]-2+mod)%mod;
            }
        }
    }
    ans=ans*f[num]%mod;
    cout<<ans<<endl;
}