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;
struct UnionFind
{
    vector<ll> par;
    vector<ll> rank;
    vector<ll> sz;
    UnionFind(ll n) : par(n), rank(n), sz(n)
    {
        for(ll i = 0; i < n; i++)
        {
            par[i] = i;
            rank[i] = 0;
            sz[i] = 1;
        }
    }
    ll root(ll x)
    {
        if(par[x] == x) return x;
        else return par[x] = root(par[x]);
    }
    void unite(ll x, ll y)
    {
        x = root(x);
        y = root(y);
        if(x == y) return;
        if(rank[x] < rank[y])
        {
            par[x] = y;
            sz[y] += sz[x];
        }
        else
        {
            par[y] = x;
            sz[x] += sz[y];
            if(rank[x] == rank[y]) rank[x]++;
        }
    }
    bool same(ll x, ll y)
    {
        return root(x) == root(y);
    }
    ll size(ll x)
    {
        x = root(x);
        return sz[x];
    }
};

int main()
{
    int n, m;
    cin >> n >> m;
    vector<int> a(m);
    vector<int> b(m);
    for(int i = 0; i < m; i++)
    {
        cin >> a[i] >> b[i];
        a[i]--;b[i]--;
    }
    int ans = 0;
    for(int i = 0; i < m; i++)
    {
        UnionFind tree(n);
        for(int j = 0; j < m; j++)
        {
            if(i == j) continue;
            tree.unite(a[j], b[j]);
        }
        if(!tree.same(a[i], b[i])) ans++;
    }
    cout << ans << endl;
}