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>

int main(){
	int n,m;
	int i,j;
	int a,b;
	int count;
	
	scanf("%d %d",&n,&m);
	
	int visited[n+1];
	int lists[n+1][n+1];
	
	for(i=0;i<=n;i++){
		visited[i] = 0;
		for(j=0;j<=n;j++){
			lists[i][j] = 0;
		}	
	}
	
	for(i=0;i<m;i++){
		scanf("%d %d",&a,&b);
		lists[a][b] = 1;
		lists[b][a] = 1;
	}
	
	visited[1] = 1;
	count =  walk(n,visited,lists,1);
	
	printf("%d\n",count);
	
	return 0;
}

int walk(int n,int visited[n+1],int lists[n+1][n+1],int k){
	int i;
	int allvisited = 1;
	int result = 0;
	
	for(i=1;i<=n;i++){
		if(visited[i]==0) allvisited = 0;
	}
	if(allvisited==1) return 1;
	
	visited[k] = 1;
	for(i=0;i<=n;i++){
		if(lists[k][i]==0) continue;
		if(visited[i]==1) continue;
		visited[i] = 1;
		result += walk(n,visited,lists,i);
		visited[i] = 0;
	}
	
	return result;
}