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++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
using namespace std;

int main() {
  int N; // <= 8
  int M; // <= 8 * (8 - 1) / 2 = 28
  cin >> N >> M;

  vector<vector<bool>> G(N, vector<bool>(N, false));
  for (int i = 0; i < M; i++) {
    int a, b;
    cin >> a >> b;
    a--; b--; // 0-indexed
    // 無向グラフ
    G.at(a).at(b) = true;
    G.at(b).at(a) = true;
  }

  // 順列用の配列データを作成する
  vector<int> v(N);
  for (int i = 0; i < N; i++) {
    v.at(i) = i;
  }

  // ソート済みのため、next_permutation で全パターンを検索する
  int ans = 0;
  do {
    // 0 から開始ではない組合せはスキップ
    if (v.at(0) != 0) {
      continue;
    }

    // 0 から開始のパターンについて、パスが全て存在するかチェック
    // 存在しなければ false とする
    bool enabling = true;
    for (int i = 0; i < (N - 1); i++) {
      int from = v.at(i);
      int to = v.at(i + 1);
      if (G.at(from).at(to) == false) {
        enabling = false;
      }
    }

    if (enabling) {
      ans++;
    }
  } while (next_permutation(v.begin(), v.end()));

  cout << ans << endl;
  return 0;
}