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 Python (3.4.3) (judge verdict: Accepted)

import sys
from collections import defaultdict
import copy

def is_connect(adj):
    is_already = defaultdict(bool)
    is_already[list(adj.keys())[0]] = True;

    queue = []
    queue.append(1)
    while len(queue) > 0:
        v = queue.pop()
        for node in adj[v]:
            if not is_already[node]:
                is_already[node] = True
                queue.append(node)

    return all([is_already[node] for node in adj.keys()])

def solve(n,m,adj,edges):
    ans = 0
    for (v1, v2) in edges:
        tmp_adj = copy.deepcopy(adj)
        tmp_adj[v1].remove(v2)
        tmp_adj[v2].remove(v1)
        if not is_connect(tmp_adj):
            ans += 1

    return ans

def main():
    s = sys.stdin.readline().rstrip()
    lst = s.split(' ')
    n = int(lst[0])
    m = int(lst[1])

    adj = defaultdict(list)
    edges = []

    for i in range(m):
        s = sys.stdin.readline().rstrip()
        lst = s.split(' ')
        a = int(lst[0])
        b = int(lst[1])
        edges.append((a,b))
        if not b in adj[a]:
            adj[a].append(b)

        if not a in adj[b]:
            adj[b].append(a)

    ans = solve(n,m,adj,edges)
    print(ans)
    return 0

if __name__ == '__main__':
    sys.exit(main())