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 C# (Mono 4.6.2.0) (judge verdict: Accepted)

//
// ABC054C
// URL : https://atcoder.jp/contests/abc054/tasks/abc054_c
// 
using System;
using System.Collections.Generic;
using System.Linq;
using System.IO;
using System.Text;
using System.IO.Pipes;

namespace AtCoder
{
    class ABC054C
    {
        public static readonly long MOD_CONST = 1000000007;//(long)Math.Pow(10, 9) + 7;


        static void Main(string[] args)
        {
            //出力高速化
            var sw = new StreamWriter(Console.OpenStandardOutput()) { AutoFlush = false };
            Console.SetOut(sw);
            Solve();
            Console.Out.Flush();
        }
        public static int N;
        public static int M;
        public static bool[,] Edges;
        public static int ans;
        public static void Solve()
        {
            {
                int[] inputArray = Console.ReadLine().Split(' ').Select(_inputStr => int.Parse(_inputStr)).ToArray();
                N = inputArray[0];
                M = inputArray[1];
            }
            Edges = new bool[N,N];
            for (int i = 0; i < M; ++i)
            {
                var inputArray = Console.ReadLine().Split(' ').Select(_inputStr => int.Parse(_inputStr)).ToArray();
                int a = inputArray[0] - 1;
                int b = inputArray[1] - 1;
                Edges[a, b] = true;
                Edges[b, a] = true;
            }

            ans = 0;
            var visited = new bool[N];
#if false
            FindPathesSub(0, ref visited);
#else
            visited[0] = true;
            ans = FindPathesSub2(0, ref visited);
#endif
            Console.WriteLine(ans);
        }

        public static void FindPathesSub(int start, ref bool[] visited)
        {
            //訪れていたら戻る
            if(visited[start] == true) { return; }
            //訪れたらチェック
            visited[start] = true;

            //全頂点訪れたら
            if (visited.All(x => x == true) == true)
            {
                //探索終了
                ++ans;//求めるパスが存在したということで、カウントアップ
                return;
            }
            //探索続行

            //次に行く頂点を探す
            var nextNodes = new List<int>(N);
            for(int next = 0; next < N; ++next)
            {
                if(Edges[start, next] == false) { continue; }//この頂点間にパスはない
                nextNodes.Add(next);
                //Console.Write (next + " ");
            }
            //Console.WriteLine();

            //次のノードに移動する
            //行くノードがないということはない、必ずどこかの頂点には移動しようとする

            //次のノードが一つだけ
            if (nextNodes.Count == 1)
            {
                //今のvisitedを使ってもう一度探索する
                FindPathesSub(nextNodes[0], ref visited);
            }
            else
            {
                //次のノードが複数
                foreach (var next in nextNodes)
                {
                    //これまでの探索結果を複製して、次のノードへ
                    var visitedCopy = new bool[N];
                    for (int i = 0; i < visited.Length; ++i) { visitedCopy[i] = visited[i]; }
                    FindPathesSub(next, ref visitedCopy);
                }
            }
        }

        /// <summary>
        /// 別回答版
        /// </summary>
        /// <param name="start"></param>
        /// <param name="visited"></param>
        public static int FindPathesSub2(int start, ref bool[] visited)
        {
            //全頂点訪れたら
            if (visited.All(x => x == true) == true)
            {
                //探索終了
                ++ans;//求めるパスが存在したということで、カウントアップ
                return 1;
            }
            //探索続行

            int ret = 0;

            //次に行く頂点を探す
            for (int next = 0; next < N; ++next)
            {
                if (Edges[start, next] == false) { continue; }//この頂点間にパスはないのでパス
                if (visited[next] == true) { continue; }//訪問済みならパス
                //訪れたとしてチェック
                visited[next] = true;
                //この頂点から先に移動
                {
                    ret += FindPathesSub2(next, ref visited);
                }
                //訪れてないとしてチェックを外すことで、もう一度この頂点から次の頂点へ探索できる
                visited[next] = false;
            }
            return ans;
        }


        //---------------------------------------------------------------------------------
#region MyLib

        /// <summary>
        /// グラフ系
        /// </summary>
        public class Node
        {
            public Node(long val){
                this.Value = val;
                this.Children = new List<Node>();
            }

            public Node(long val, List<Node> children)
            {
                this.Value = val;
                this.Children = children;
            }

            public long Value { get; set; }
            public List<Node> Children { get; set; }
        }

        /// <summary>
        /// Clamping
        /// </summary>
        /// <typeparam name="T">比較型</typeparam>
        /// <param name="x">clampしたい値</param>
        /// <param name="min">最小値</param>
        /// <param name="max">最大値</param>
        /// <returns>clamp済みの値</returns>
        public static T Clamp<T>(T x, T min, T max) where T : IComparable<T>
        {
            if (x.CompareTo(min) < 0) { return min; }
            else if (x.CompareTo(max) > 0) { return max; }
            else return x;
            //return Math.Max(Math.Min(x, max), min);
        }

        /// <summary>
        /// 指定された底log_baseでの, 対数を計算
        /// </summary>
        /// <param name="x"></param>
        /// <param name="log_base">指定の底</param>
        /// <returns></returns>
        public static double LogBase(double x, double log_base)
        {
            //log_b(N) = log_a(N) / log_a(B)より
            return Math.Log(x) / Math.Log(log_base);
        }

        /// <summary>
        /// 初項a, 公差d, 項数nの数列の和
        /// </summary>
        /// <param name="n">求めたいところ</param>
        /// <returns>初項からxまで足した値</returns>
        static long SumSeries(long a, long d, long n)
        {
            return (2 * a + (n - 1) * d) * n / 2;
        }
        /// <summary>
        /// 累積和の配列を計算する
        /// Sum[ a[0], a[i] )がs[i+1]に入っている
        /// s[0]  : 0                //一切和を取らない場合
        /// s[1]  : ls[0]            //a[0]   + s[0] = a[0]
        /// s[2]  : ls[1] + s[1]     //a[1]   + a[0]
        /// ...
        /// s[N]  : ls[N-1] + s[N-1] //a[N-1] + ... + a[0]
        /// s[N+1]: ls[N] + s[N]     //a[N]   + ... + a[0]
        /// 
        /// 例: [1, 7)の和(a[1] + ... + a[6])
        /// s[1] = a[0] //[0, 1)
        /// s[7] = a[0] + a[1] + ... + a[6] //[0, 7)
        /// よってs[7] - s[1]);
        /// </summary>
        /// <param name="ls">計算したい配列</param>
        /// <returns>累積和配列</returns>
        static List<long> AggregateSumList(List<long> a)
        {
            var s = new List<long>(a.Count + 1);
            if (a.Count == 0)
            {
                return s;
            }
            s.Add(0);//s[0] = 0
            for (int i = 0; i < a.Count; ++i)
            {
                s.Add(a[i] + s[i]);
            }
            return s;
        }


        /// <summary>
        /// 与えられた数が素数かどうかを計算する
        /// </summary>
        /// <param name="num">対象数</param>
        /// <returns>true:素数, false:素数ではない(合成数)</returns>
        public static bool isPrime(int num)
        {
            if (num < 2) return false;
            else if (num == 2) return true;
            else if (num % 2 == 0) return false;

            double sqrtNum = Math.Sqrt(num);
            for (int i = 3; i <= sqrtNum; i += 2)
            {
                if (num % i == 0)
                {
                    return false;
                }
            }
            return true;
        }

        /// <summary>
        /// 順列mPnを計算する
        /// </summary>
        /// <param name="m"></param>
        /// <param name="n"></param>
        /// <returns></returns>
        public static long Permutation(long m, long n)
        {
            long ret = 1;
            for (var i = 0; i < n; ++i)
            {
                ret *= m - i;
            }
            return ret;
        }
        /// <summary>
        /// 組み合わせmCnを計算する
        /// </summary>
        /// <param name="m"></param>
        /// <param name="n"></param>
        /// <returns></returns>
        public static long Combination(long m, long n)
        {
            if (m == n || n == 0)
            {
                return 1;
            }
            long ans = 1;
            for (var i = 1; i <= n; ++i)
            {
                ans = ans * (m - i + 1) / i;
            }
            return ans;
        }

        static long Gcd(long m, long n)
        {
            if (n == 0) return m;
            return Gcd(n, m % n);
        }

        static long Lcm(long m, long n)
        {
            var gcd = Gcd(m, n);
            //return (m * n) / gcd;
            return m / gcd * n;//桁あふれ対策
        }
#endregion MyLib
    }
}