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 Kotlin (1.0.0) (judge verdict: Accepted)

import java.util.*
class SearchInfo(val nodeNumber: Int, val visitedNodes: Set<Int>)
fun main(args : Array<String>) {
    val (N, M) = readLine()!!.split(" ").map { it.toInt() }
    val graph = Array(N, { Array(N, { false }) })

    (1..M).forEach {
        val (a, b) = readLine()!!.split(" ").map { it.toInt() }
        graph[a-1][b-1] = true
        graph[b-1][a-1] = true
    }

    val queue : Queue<SearchInfo>  = ArrayDeque()

    graph[0].forEachIndexed { nodeNum, exists ->
        if (exists) {
            queue.offer(SearchInfo(nodeNum, setOf(0, nodeNum)))
        }
    }

    var count = 0
    while (queue.isNotEmpty()) {
        val searchInfo = queue.poll()
        if (searchInfo.visitedNodes.size == N) {
            count++
        } else {
            val currentNodeNum = searchInfo.nodeNumber
            graph[currentNodeNum].forEachIndexed{ node, exists ->
                if (exists && !searchInfo.visitedNodes.contains(node)) {
                    //println(searchInfo.visitedNodes.plus(node).toString())
                    queue.offer(SearchInfo(node, searchInfo.visitedNodes.plus(node)))
                }
            }
        }
    }

    println(count)
}