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: Wrong Answer)

#include "bits/stdc++.h"
#define rep(i,n) for (int i=0; i<(n); ++i)
using namespace std;
using ll =long long;
using P =pair<int,int>;

int main(){
    int n,m;
    cin >> n >> m;
    vector <int> a(m),b(m);
    vector <vector <int>> ma(n);
    rep(i,m){
        cin >> a[i] >> b[i];
        ma[a[i]-1].push_back(b[i]-1);
        ma[b[i]-1].push_back(a[i]-1);
         
    }

    queue <int> q;
    int qq;
    int ans=0;
    vector <int> to(n,0),to2(n,0);
    to2[0]=1;
    int k=0;
    int l;
    rep(i,n){
        k=1;
        to=to2;
        q.push(0);
        while(q.size()!=0){
            qq=q.front();
            //cout << qq << endl;
            q.pop();
            l=ma[qq].size();
            rep(j,l){
                if(a[i]==qq+1 and b[i]==ma[qq][j]+1){
                    k=k;
                }
                else if(b[i]==qq+1 and a[i]==ma[qq][j]+1){
                    k=k;
                }
                else{
                    if(to[ma[qq][j]]==0){
                        q.push(ma[qq][j]);
                        to[ma[qq][j]]=1;
                        ++k;
                    }
                }
            }
        }
        //cout << k << endl;
        if(k!=n){
            ++ans;
        }
    }

    cout << ans << endl;


    

    return 0;
}