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 <iostream>
#include <string>
#include <vector>
#include <algorithm>
#include <map>
#include <set>
#include <functional>
#include <limits>
#include <queue>
#include <iomanip>
#include <deque>
#include <stack>
#include <cstdio>
#include <cmath>
#include <complex.h>
#include <cstdio>
#include <random>
#include <chrono>
#include <numeric>
#include <cassert>
#include <iso646.h>
#include <exception>
#include <stdexcept>

using namespace std;

typedef long long i64;

vector<int> adj[60];
int edges[2500][2];

bool visited[60];

void dfs(const int v, const int excluded_edge)
{
    visited[v] = true;

    for (int adj_v : adj[v])
    {
        bool bad_edge = visited[adj_v];
        bad_edge |= edges[excluded_edge][0] == v && edges[excluded_edge][1] == adj_v;
        bad_edge |= edges[excluded_edge][1] == v && edges[excluded_edge][0] == adj_v;

        if (!bad_edge)
            dfs(adj_v, excluded_edge);
    }
}

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

    int n, m;
    cin >> n >> m;

    for (int i = 0; i < m; i++)
    {
        int u, v;
        cin >> u >> v;

        adj[u].push_back(v);
        adj[v].push_back(u);

        edges[i][0] = u;
        edges[i][1] = v;
    }

    int bridges = 0;
    for (int i = 0; i < m; i++)
    {
        dfs(1, i);

        bool exists_not_visited = false;
        for (int v = 1; v <= n; v++)
        {
            exists_not_visited |= !visited[v];
            visited[v] = false;
        }

        bridges += exists_not_visited;
    }

    cout << bridges;
}