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 (GCC 5.4.1) (judge verdict: Accepted)

#include<stdio.h>

#define N 8
#define M (N*(N-1)/2)

int n, m, a[M], b[M];

int search(int noArray, int *isStoraged, int *seq, int *cnt){
  if(noArray==n){
//    for(int j=0; j<n; j++) { printf("%d",seq[j]); } printf("\n");
    *cnt = *cnt + 1;
  }
  else{
    for(int i=0; i<m; i++){
      if(a[i]==seq[noArray-1] && isStoraged[b[i]-1]==0){
        isStoraged[b[i]-1] = 1;
        seq[noArray] = b[i];
        search(noArray+1, isStoraged, seq, cnt);
        isStoraged[b[i]-1] = 0;
      }
      if(b[i]==seq[noArray-1] && isStoraged[a[i]-1]==0){
         isStoraged[a[i]-1] = 1;
         seq[noArray] = a[i];
         search(noArray+1, isStoraged, seq, cnt);
         isStoraged[a[i]-1] = 0;
      }
    }
  }
  return 0;
}

int main(){
  scanf("%d %d", &n, &m);
  for(int i=0; i<m; i++){
    scanf("%d %d", &a[i], &b[i]);
  }
  
  int isPass[N]={0}, orderPassed[N]={0}, cnt=0;
  isPass[0]=1;  // node 1 is Passed initially
  orderPassed[0]=1; //  node 1 is Storaged
  search(1, isPass, orderPassed, &cnt);
  
  printf("%d",cnt);
  
  return 0;
}