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++ (GCC 9.2.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
#define rep(i,n) for (int i = 0; i < n; i++)
#define rep2(i,a,n) for (int i = a; i < n; i++)
using namespace std;
int N, M;
vector<pair<int, int>> P;

struct UnionFind {
    vector<int> par;

    UnionFind(int N): par(N) {
        rep(i, N) par[i] = i;
    }

    int root(int x) {
        if (par[x] == x) return x;
        return par[x] = root(par[x]);
    }

    bool same(int x, int y) {
        int rx = root(x);
        int ry = root(y);
        return rx == ry;
    }

    void unite(int x, int y) {
        int rx = root(x);
        int ry = root(y);

        if (same(rx, ry)) return;

        par[rx] = ry;
    }

};


void init() {
    cin >> N >> M;

    rep(i, M) {
        int a, b; cin >> a >> b;
        a--; b--;
        P.push_back({a, b});
    }

}

void solve() {
    int res =0;
    rep(de, M) {
        bool flag = false;
        UnionFind uf(N);
        rep(i, M) {
            if (i == de) continue;
            uf.unite(P[i].first, P[i].second);
        }
        int stad = 0;
        for (int i = 1; i < N; i++) {
            if (!uf.same(stad, i)) flag = true;
        }

        if (flag) res++;
    }

    printf("%d\n", res);
}

int main() {
    init();
    solve();
    return 0;
}