Problem:
Score : 300 points

Problem Statement

You are given an undirected connected graph with N vertices and M edges that does not contain self-loops and double edges.

The i-th edge (1 \leq i \leq M) connects Vertex a_i and Vertex b_i.

An edge whose removal disconnects the graph is called a bridge.

Find the number of the edges that are bridges among the M edges.

Notes

A self-loop is an edge i such that a_i=b_i (1 \leq i \leq M).

Double edges are a pair of edges i,j such that a_i=a_j and b_i=b_j (1 \leq i<j \leq M).

An undirected graph is said to be connected when there exists a path between every pair of vertices.

Constraints

2 \leq N \leq 50

N-1 \leq M \leq min(N(N−1)⁄2,50)

1 \leq a_i<b_i \leq N

The given graph does not contain self-loops and double edges.

The given graph is connected.

Input

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 edges that are bridges among the M edges.

Sample Input 1

7 7
1 3
2 7
3 4
4 5
4 6
5 6
6 7

Sample Output 1

4

The figure below shows the given graph:

The edges shown in red are bridges. There are four of them.

Sample Input 2

3 3
1 2
1 3
2 3

Sample Output 2

0

It is possible that there is no bridge.

Sample Input 3

6 5
1 2
2 3
3 4
4 5
5 6

Sample Output 3

5

It is possible that every edge is a bridge.

Solution in C++ (GCC 5.4.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
using namespace std;
vector<int> s[55];
int book[55]={};
bool dfs(int from,int goal,int x)
{
	bool is=true;
	int i,q;
	for(i=0;i<s[from].size();i++)
	{
		q=s[from][i];
		if(q==goal&&from!=x)
		{
			return false;
		}
		if(book[q])
		{
			continue;
		}
		book[q]=1;
		is&=dfs(q,goal,x);
		book[q]=0;
	}
	if(is)
		return true;
	return false;
}
int main(){
	int n,m,a[55],b[55],times[55]={},i,j,k,l,q,p,sum=0;
	cin>>n>>m;
	for(i=0;i<m;i++)
	{
		cin>>a[i]>>b[i];
		s[a[i]].push_back(b[i]);
		s[b[i]].push_back(a[i]);
	}
	for(i=0;i<m;i++)
	{
		for(j=0;j<55;j++)
		{
			book[j]=0;
		}
		book[a[i]]=1;
		book[b[i]]=1;
		if(s[a[i]].size()<s[b[i]].size())
		{
			if(dfs(a[i],b[i],a[i]))
			{
				sum++;
			}
		}
		else
		{
			if(dfs(b[i],a[i],b[i]))
			{
				sum++;
			}
		}
	}
	cout<<sum<<endl;
	return 0;
}