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)

class Unionfind:
  def __init__(self,n):
     self.Tree=[i for i in range(n)]

  def find(self,x):
    if x==self.Tree[x]:
      return x
    else:
      root=self.find(self.Tree[x])
      self.Tree[x]=root
      return root

  def unite(self,x,y):
    s1=self.find(x)
    s2=self.find(y)
    if s1==s2:
      pass
    else:
      self.Tree[s1]=s2
      
N,M=map(int,input().split())
Uf=Unionfind(N)
Flag=0
B=0
A=[]
Edge=[[int(i) for i in input().split()] for i in range(M)]
for m in range(M):
  Edge_copy=list(Edge)
  del Edge_copy[m]
  for e in Edge_copy:
    Uf.unite(e[0]-1,e[1]-1)
  for a in range(N):
    A.append(Uf.find(a))
  for n in range(N-1):
    if Uf.find(n)!=Uf.find(n+1):
      Flag=1
      break
  if Flag==1:
    B+=1
  Flag=0
  Uf=Unionfind(N)
  A=[]
print(B)