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;
typedef long long ll;
typedef pair<int, int> pint;
const int dx[] = {-1, 0, 0, 1, -1, -1, 1, 1};
const int dy[] = {0, -1, 1, 0, -1, 1, -1, 1};

vector<int> g[50];
vector<pint> edges;
bool seen[50];
int main() {
  int n, m;
  cin >> n >> m;
  for (int i = 0; i < m; i++) {
    int a, b;
    cin >> a >> b;
    a--;
    b--;
    g[a].push_back(b);
    g[b].push_back(a);
    pint e(a, b);
    edges.push_back(e);
  }
  int result = 0;
  for (auto&& edge : edges) {
    for (int i = 0; i < n; i++) seen[i] = false;
    queue<int> todo;
    todo.push(0);
    while (!todo.empty()) {
      int p = todo.front();
      todo.pop();
      seen[p] = true;
      for (auto&& e : g[p]) {
        if ((p == edge.first && e == edge.second) ||
            (p == edge.second && e == edge.first))
          continue;
        if (!seen[e]) todo.push(e);
      }
    }
    for (int i = 0; i < n; i++) {
      if (!seen[i]) {
        result++;
        break;
      }
    }
  }
  cout << result << endl;
}