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;

int n, m;
const int MAX = 55;
bool G[MAX][MAX], used[MAX];
int A[MAX], B[MAX];

void dfs(int now){
    used[now] = true;
    for(int i = 0; i < n; i++){
        if(!used[i] && G[now][i]){
            dfs(i);
        }
    }
    return;
}

int main(){
    cin.tie(0);
    ios::sync_with_stdio(false);

    cin >> n >> m;
    for(int i = 0; i < m; i++){
        cin >> A[i] >> B[i];
        A[i]--;
        B[i]--;
        G[A[i]][B[i]] = true;
        G[B[i]][A[i]] = true;
    }

    int ans = 0;
    for(int i = 0; i < m; i++){
        G[A[i]][B[i]] = false;
        G[B[i]][A[i]] = false;
    
        for(int j = 0; j < n; j++){
            used[j] = false;
        }

        dfs(0);

        bool f = false;
        for(int j = 0; j < n; j++){
            if(used[j] == false) f = true;
        }

        if(f) ans++;

        G[A[i]][B[i]] = true;
        G[B[i]][A[i]] = true;
    }

    cout << ans << endl;


    return 0;
}