Problem:
Score : 300 points

Problem Statement

You are given an undirected unweighted graph with N vertices and M edges that contains neither self-loops nor double edges.

Here, a self-loop is an edge where a_i = b_i (1≤i≤M), and double edges are two edges where (a_i,b_i)=(a_j,b_j) or (a_i,b_i)=(b_j,a_j) (1≤i<j≤M).

How many different paths start from vertex 1 and visit all the vertices exactly once?

Here, the endpoints of a path are considered visited.

For example, let us assume that the following undirected graph shown in Figure 1 is given.

Figure 1: an example of an undirected graph

The following path shown in Figure 2 satisfies the condition.

Figure 2: an example of a path that satisfies the condition

However, the following path shown in Figure 3 does not satisfy the condition, because it does not visit all the vertices.

Figure 3: an example of a path that does not satisfy the condition

Neither the following path shown in Figure 4, because it does not start from vertex 1.

Figure 4: another example of a path that does not satisfy the condition

Constraints

2≦N≦8

0≦M≦N(N-1)/2

1≦a_i<b_i≦N

The given graph contains neither self-loops nor double edges.

Input

The input is given from Standard Input in the following format:

N M
a_1 b_1
a_2 b_2
:
a_M b_M

Output

Print the number of the different paths that start from vertex 1 and visit all the vertices exactly once.

Sample Input 1

3 3
1 2
1 3
2 3

Sample Output 1

2

The given graph is shown in the following figure:

The following two paths satisfy the condition:

Sample Input 2

7 7
1 3
2 7
3 4
4 5
4 6
5 6
6 7

Sample Output 2

1

This test case is the same as the one described in the problem statement.

Solution in Scala (2.11.7) (judge verdict: Accepted)

import scala.io.{ StdIn => in}
import scala.annotation.tailrec

object Main extends App {
  val Array(n, m) = in.readLine.split(" ").map(_.toInt)
  val graph = Array.ofDim[Boolean](n, n)

  Iterator.continually(in.readLine)
    .take(m)
    .map(_.split(" ").map(_.toInt - 1))
    .foreach(e => {
      graph(e(0))(e(1)) = true
      graph(e(1))(e(0)) = true
    })

  def solve():Int = {
    var cnt = 0
    val p = Array.ofDim[Int](n)

    def evalute():Unit = {
      // println(p.mkString(" "))
      @tailrec
      def rec(idx:Int):Boolean = {
        val next = idx+1
        if(next == n) true
        else if(graph(p(idx))(p(next))) rec(next)
        else false
      }

      if(rec(0)) cnt += 1
    }

    def dfs(pos:Int, mask:Int):Unit = {
      if(pos == n) evalute
      else {
        for(i <- 0 until n if (mask & (1<<i)) > 0) {
          p(pos) = i
          dfs(pos+1, mask ^ (1<<i))
        }
      }
    }

    dfs(1, ((1<<n)-1) ^ 1<<0)
    cnt
  }

  println(solve)
}