Codeforces Round #369 (Div. 2) -- D. Directed Roads （DFS找环）

大体题意：

给你一个有向图，可能会有环，你的操作是反向一条路，求得使得图中没有环所有方案数？

思路：

假如图中没有环的话，有n条边，答案就是2^n

如果有个m边的环，间接法考虑，总方案是2^m,  操作不改变的环的方法有2种，  什么也不动，和  每一个边都反向。

所有破坏环有2^m - 2种操作！

所以这个问题转换为 我们求出图中有几个环，求出操作数，在求剩下不成环的操作！

详细见代码：

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxn = 200000 + 10;
const int mod = 1e9+7;
int sum = 0,vis[maxn],cnt = 0,dis[maxn],a[maxn];
ll ans = 1ll;
int POW[maxn];
ll num = 0ll;
void init(){
POW[0] = 1;
for (int i = 1; i < maxn; ++i)POW[i] = (POW[i-1] % mod * 2 % mod)%mod;
}
void dfs(int cur,int val){
vis[val] = cnt;
dis[val] = cur;
if (vis[a[val]] != 0 && vis[a[val]] == cnt){
sum = (cur - dis[a[val]] + 1);
num += sum;
ans = (ans % mod * (POW[sum]-2) % mod) % mod;
}
else if (!vis[a[val]])dfs(cur+1,a[val]);
}
int main(){
int n;
init();
scanf("%d",&n);
for (int i = 1; i <= n; ++i){
scanf("%d",&a[i]);
}
for (int i = 1; i <= n; ++i){
if (!vis[i]){
++cnt;
dfs(0,i);
}
}
ans = (ans % mod * POW[n-num] % mod) % mod;
printf("%I64d\n",ans);
return 0;
}

[align=center]D. Directed Roads[/align]

time limit per test
2 seconds

memory limit per test
256 megabytes

input
standard input

output
standard output

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.

Examples

Input

3
2 3 1

Output

6

Input

4
2 1 1 1

Output

8

Input

5
2 4 2 5 3

Output

28

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

,


,


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.