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++14 (GCC 5.4.1) (judge verdict: Accepted)

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

void bfs(ll node, vector<bool>& seen, const vector<pair<ll, ll>>& edge, const pair<ll, ll>& remove_edge)
{
    if (seen[node - 1]) {
        return;
    }

    seen[node - 1] = true;
    for (auto&& e : edge) {
        if ((e.first != node && e.second != node)
            || (e == remove_edge)) {
            continue;
        }
        if (e.first == node) {
            bfs(e.second, seen, edge, remove_edge);
        }
        if (e.second == node) {
            bfs(e.first, seen, edge, remove_edge);
        }
    }
}

void main_()
{
    ll N, M;
    cin >> N >> M;
    vector<pair<ll, ll>> edge;
    for (ll i = 0; i < M; i++) {
        ll a, b;
        cin >> a >> b;
        edge.emplace_back(a, b);
    }

    ll ans = 0;
    for (ll i = 0; i < M; i++) {
        vector<bool> seen(N);
        bfs(edge.front().first, seen, edge, edge[i]);
        if (!all_of(begin(seen), end(seen), [](bool s) { return s; })) {
            ans++;
        }
    }
    cout << ans << endl;
}
int main()
{
    cin.tie(nullptr);
    cout.tie(nullptr);
    ios_base::sync_with_stdio(false);
    main_();
    return 0;
}