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)

# pythonでスタック、キューを実現するためにdequeをimport
from collections import deque

N, M = map(int, input().split())

# 辺集合EとグラフGの構築
G = [[] for _ in range(N)]
E = []
for _ in range(M):
    a, b = map(int, input().split())
    # 入力は1-indexedだがこれを扱いやすいように0-indexedに変更
    a -= 1
    b -= 1
    E.append((a, b))
    G[a].append(b)
    G[b].append(a)
# print(G)

# 橋を数えるための変数cntを用意
cnt = 0
for edge_to_remove in E:
    # 橋であるか判定したい辺を取り除いてみる
    i, j = edge_to_remove
    G[i].remove(j)
    G[j].remove(i)

    # 深さ優先探索
    seen = [False] * N
    todo = deque()
    # 適当な頂点から深さ優先探索をスタート（ここでは頂点0をスタートに選んだ）
    todo.append(0)
    seen[0] = True
    seen_nodes_num = 0
    while todo:
        here = todo.pop()
        seen_nodes_num += 1
        for neighbor in G[here]:
            if seen[neighbor] == False:
                seen[neighbor] = True
                todo.append(neighbor)
    # print(seen)
    # もし訪問できていない頂点があれば連結でないというなので今着目している辺は橋であった！
    if seen_nodes_num != N:
        cnt += 1

    # 次の辺について調べるために今取り除いた辺を再びGに戻しておく
    G[i].append(j)
    G[j].append(i)
print(cnt)