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, size) :
        self.parent = list(range(size))
        self.height = [0] * size
        self.size = [1] * size
        self.component = size

    def root(self, index) :
        if self.parent[index] == index :  # 根の場合
            return index
        rootIndex = self.root(self.parent[index])  # 葉の場合親の根を取得
        self.parent[index] = rootIndex  # 親の付け直し
        return rootIndex

    def union(self, index1, index2) :  # 結合
        root1 = self.root(index1)
        root2 = self.root(index2)

        if root1 == root2 :  # 連結されている場合
            return

        self.component -= 1  # 連結成分を減らす

        if self.height[root1] < self.height[root2] :
            self.parent[root1] = root2  # root2に結合
            self.size[root2] += self.size[root1]
        else :
            self.parent[root2] = root1  # root1に結合
            self.size[root1] += self.size[root2]
            if self.height[root1] == self.height[root2] :
                self.height[root1] += 1
        return

    def isSameRoot(self, index1, index2) :
        return self.root(index1) == self.root(index2)

    def sizeOfSameRoot(self, index) :
        return self.size[self.root(index)]

    def getComponent(self) :
        return self.component

N, M = map(int, input().split())
AB = [tuple(map(lambda a: int(a) - 1, input().split())) for _ in range(M)]

ans = 0
for i in range(M):
    tree = UnionFind(N)
    for a, b in (AB[:i] + AB[i + 1:]):
        tree.union(a, b)
    if tree.getComponent() > 1:
        ans += 1
print(ans)