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>
#define ll long long
#define P pair<int,int>
#define MAX_N 51
using namespace std;

bool G[MAX_N][MAX_N];
bool visited[MAX_N];
int n,m;

void dfs(int u){
    visited[u] = true;
    for(int v = 0; v < n; v++){
        if( !G[u][v] ) continue;
        if( visited[v] )continue;
        dfs(v);
    }
}

int main(){
    ios_base::sync_with_stdio(0); cin.tie(0);
    cin>>n>>m;
    int b1[m];
    int b2[m];
    for(int i = 0;i<m;i++){
        cin>> b1[i] >>b2[i];
        b1[i]--;b2[i]--;
        G[b1[i]][b2[i]] = G[b2[i]][b1[i]] = true;
        
    }
    ll ans = 0;
    for(int i = 0; i<m;i++){
        G[b1[i]] [b2[i]] = G[b2[i]] [b1[i]] = false;
        for(int i=0; i<n;++i) visited[i] = false;

        dfs(0);

        bool can = true;
        for(int j =0; j<n; ++j){
            if( !visited[j] ) {
                can = false;
                break;
            }
        }
        if( !can ) ans++;
        G[b1[i]] [b2[i]] = G[b2[i]] [b1[i]] =true;
    }
    cout<< ans <<endl;
}