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 java.util.Scanner

/**
 * Created by sonetsuyoshi on 2015/12/16.
 */

object Main extends App {
  val sc = new Scanner(System.in)

  val n = sc.nextInt
  val m = sc.nextInt

  val connected = Array.ofDim[Boolean](n,n)
  val visited = Array.ofDim[Boolean](n)

  val ids = 0.until(n)

  for(i <- 0 until m) {
    val a = sc.nextInt - 1
    val b = sc.nextInt - 1
    connected(a)(b) = true
    connected(b)(a) = true
  }

  run()

  def run(): Unit = {
    visited(0) = true
    val ans = search(0)
    println(ans)
  }

  def search(id: Int): Int = {
    val adjs = ids.filter(i => connected(id)(i) && !visited(i))

    if(adjs.size == 0) {
      if(visited.forall(x => x)) 1
      else 0
    } else {
      adjs.foldLeft(0){(s, i) => {
        visited(i) = true
        val s1 = s + search(i)
        visited(i) = false
        s1
      }}
    }
  }
}