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;
#define rep(i,n) for(int i=0;i<n;++i)
#define rep2(i,i0,n) for(int i=i0;i<n;++i)

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

    int ans = 0;
    vector<bool> seen(n, false);
    queue<int> que;

    rep(i,m) {
        fill(seen.begin(),seen.end(),false);
        int seen_cnt = 0;
        que.push(0);
        seen[0] = true;
        ++seen_cnt;

        while(!que.empty()) {
            int cv = que.front();
            que.pop();
            for(auto nv:g[cv]) {
                if(seen[nv]) continue;
                if(cv == ab[i].first && nv == ab[i].second)
                    continue;
                if(cv == ab[i].second && nv == ab[i].first)
                    continue;
                que.push(nv);
                seen[nv] = true;
                ++seen_cnt;
            }
        }

        if(seen_cnt != n) ++ans;
    }

    cout << ans << endl;
    return 0;
}