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 int long long
#define FOR(i, j, k) for(int i = j; i < k; ++i)
#define rep(i, j) FOR(i, 0, j)
#define INF (1 << 30)
#define fi first
#define se second

typedef unsigned long long ull;
typedef pair<int, int> P;
typedef pair<P, int> Pi;
typedef pair<P, P> PP;

const int MOD = 1e9 + 7;
const int dy[]={0, 0, 1, -1};
const int dx[]={1, -1, 0, 0};

template <class T> void chmin(T& a, const T& b) { a = min(a, b); }
template <class T> void chmax(T& a, const T& b) { a = max(a, b); }

P memo[50];
int dist[50][50];
int res;
int n, m;

void Warshall_Floyd() {
	for(int k = 0; k < n; ++k) for(int i = 0; i < n; ++i) for(int j = 0; j < n; ++j) {
		dist[i][j] = min(dist[i][j], dist[i][k] + dist[k][j]);
	}
}

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

    cin >> n >> m;
    rep(i, m) {
        int a, b;
        cin >> a >> b;
        --a; --b;
        memo[i] = P(a, b);
        dist[a][b] = dist[b][a] = 0;
    }
    rep(i, m) {
        int a = memo[i].first, b = memo[i].second;
        rep(j, n) rep(k, n) {
            dist[j][k] = (j != k) ? INF : 0;
        }
        rep(j, m) {
            int aa = memo[j].first, bb = memo[j].second;
            dist[aa][bb] = dist[bb][aa] = 0;
        }
        dist[a][b] = dist[b][a] = INF;
        Warshall_Floyd();
        if(dist[a][b] != 0) ++res;
        dist[a][b] = dist[b][a] = 0;
    }
    cout << res << endl;
    return 0;
}