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 <iostream>
#include <algorithm>
using namespace std;
//擒贼先擒王，找boss
//断桥问题 只要断开一条边boss就肯定有不相同的了
int N, M, a[55], b[55], par[55];
int root(int x)
{
    if (x == par[x]) return x;   //连接到他自己本身那么他自己就是boss
    return par[x] = root(par[x]);  //这里又开始递归，找更强的boss
}
void unite(int x, int y)
{
    x = root(x);
    y = root(y);
    par[x] = y;
}
int main()
{
    cin >> N >> M;
    for (int i = 0; i < M; i++) cin >> a[i] >> b[i], a[i]--, b[i]--;
    int ret = 0;
    //对所有边进行遍历检索
    for (int i = 0; i < M; i++)
    {
        //一开始默认每个点自己都是boss
        for (int j = 0; j < N; j++) par[j] = j;
        //对剩下其他边处理（相当于把i所在的边空余出来，i边相当于断掉了）
        for (int j = 0; j < M; j++)
        {
            if (i != j) unite(a[j], b[j]); //找两个点间的boss
        }
        bool f = false;
        //因为对于点两两作比较所以是1->n-1
        for (int j = 1; j < N; j++)
        {
            if (root(j - 1) != root(j)) f = true;   //如果两点间的boss不同那么存在断点，符合题意，计数器加一
        }
        if (f) ret++;
    }
    cout << ret << endl;
    return 0;
}