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 9.2.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
#define ll long long int
#define mod_ceil(a, b) (a % b) ? a / b + 1 : a / b
using namespace std;

int main()
{
    ios::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);

    ll n, m, a, b, ans = 0;
    cin >> n >> m;
    vector<ll> G[n];
    pair<ll, ll> E[m];
    vector<bool> B(m, false);
    for (ll i = 0; i < m; i++)
    {
        cin >> a >> b;
        G[a - 1].push_back(b - 1);
        G[b - 1].push_back(a - 1);
        E[i] = {a - 1, b - 1};
    }
    for (ll i = 0; i < m; i++)
    {
        ll l = E[i].first, r = E[i].second;
        for (auto it = G[l].begin(); it != G[l].end(); it++)
        {
            if (*it == r)
            {
                G[l].erase(it);
                break;
            }
        }
        for (auto it = G[r].begin(); it != G[r].end(); it++)
        {
            if (*it == l)
            {
                G[r].erase(it);
                break;
            }
        }
        vector<bool> vis(n, false);
        queue<ll> Q;
        Q.push(l);
        while (!Q.empty())
        {
            ll top = Q.front();
            Q.pop();
            if (vis[top])
                continue;
            vis[top] = true;
            for (ll j = 0; j < G[top].size(); j++)
            {
                if (!vis[G[top][j]])
                {
                    Q.push(G[top][j]);
                }
                if (G[top][j] == r)
                {
                    B[i] = true;
                }
            }
        }
        G[l].push_back(r);
        G[r].push_back(l);
    }
    cout << count(B.begin(), B.end(), false) << endl;
}