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 <bits/stdc++.h>
using namespace std;

#define rep(i, n) for (int i = 0; i < n; i++)

#define limit 50

bool graph[limit][limit] = {false};
bool pass[limit] = {false};

int ans = 0;

void dfs(int v, int n) {
    pass[v] = true;

    for (int v2 = 0; v2 < n; v2++) {
        if (!graph[v][v2])
            continue;
        if (pass[v2])
            continue;

        dfs(v2, n);
    }
}

//dfs問題

int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);

    int n, m;
    cin >> n >> m;

    vector<int> a(m), b(m);

    rep(i, m) {
        cin >> a[i] >> b[i];

        a[i]--;
        b[i]--;

        graph[a[i]][b[i]] = true;
        graph[b[i]][a[i]] = true;
    }

    int ans = 0;
    rep(i, m) {
        graph[a[i]][b[i]] = false;
        graph[b[i]][a[i]] = false;

        rep(j, n) pass[j] = false;

        dfs(0, n);

        bool bb = false;
        rep(j, n) if (pass[j] == false) bb = true;

        if (bb) {
            ans++;
        }

        graph[a[i]][b[i]] = true;
        graph[b[i]][a[i]] = true;
    }

    cout << ans << endl;
}