Directed Roads
Contest: Codeforces Round 369 (Div. 2)
Difficulty rating: 1900
Tags: combinatorics, dfs and similar, graphs, math

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 format:
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 format:
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.

Constraints:
Time limit: 2.0 s
Memory limit: 256.0 MB

Note:
Consider the first sample case. There are 3 towns and 3 roads. The towns are numbered from 1 to 3 and the roads are $$1 \rightarrow 2$$, $$2\rightarrow3$$, $$\overset{3}{\rightarrow}1$$ initially. Number the roads 1 to 3 in this order.

The sets of roads that ZS the Coder can flip (to make them not confusing) are {1}, {2}, {3}, {1, 2}, {1, 3}, {2, 3}. Note that the empty set is invalid because if no roads are flipped, then towns 1, 2, 3 is form a directed cycle, so it is confusing. Similarly, flipping all roads is confusing too. Thus, there are a total of 6 possible sets ZS the Coder can flip.

The sample image shows all possible ways of orienting the roads from the first sample such that the network is not confusing.

Example input:
3
2 3 1

Example output:
6

Example input:
4
2 1 1 1

Example output:
8

Editorial:
We want to find the number of ways to assign directions to the edges of the graph so that it is acyclic, i.e. contains no cycles. First, we investigate what the graph looks like. It can be easily seen that the graph consists of several connected components, where each component is either a path or a cycle with some paths connected to some of the vertices. We'll solve the problem for each component and multiply the result for each component. If the component is a path, then however we orient the edges we won't form a cycle. So, if there are x edges in the component, there are 2x ways to orient the edges. Next, if the component has a cycle, then we can orient the remaining edges in any way (it will not affect the existence of cycles), then for the x edges on the cycle, we have 2x - 2 ways to orient them. This is because there are a total of 2x ways to orient the edges, but only 2 ways gives a directed cycle. (fix the direction of one of the edges, then the directions of the other edges are uniquely determined to form a directed cycle) Thus, if there are t edges not on the cycle and x edges on the cycle, the total number of ways for this component is 2t(2x - 2). Finally, to compute the final answer we multiply the answers for each component. Computing the powers of 2 can be done by a simple precomputation or using binary exponentiation. (the former works because the number of vertices is small) Finding the cycles can be easily done with a dfs.
Time Complexity : O(n)