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: Wrong Answer)

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

#define MAX_V  8

vector<int> G[ MAX_V ]; 
int visited[ MAX_V ] ;
int nt[ MAX_V ] ;
int cnt = 0;
int res = 0;
int V,E;
int s,t;

int next(int u){
    for(int j = nt[u];  j < G[u].size(); j++){
        int v = G[u][j];
        if( visited[v] != 1 ){
            nt[u] = j;
            return v;
        }
    }
    return -1;
}

int dfs(){
    stack<int>st;
    st.push(1);
    visited[1] = 1;
    cnt++;

    while( !st.empty() ){
        if(cnt == V)res++;
        int u = st.top();
        int v = next(u);
        if(v != -1){
            
            st.push(v);
            visited[v] = 1;
            cnt++;
        }else{
            st.pop();
            visited[u] = 2;
            cnt--;
        }
    }
}

int main(){
    cin>>V>>E;
    for(int i =0; i < E; i++){
        cin>>s>> t;
        G[s].push_back(t);
        G[t].push_back(s);
    }
    fill(nt,nt + V,1);
    fill(visited, visited + V, 0);
    dfs();
    printf("%d¥n", res);
}