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 <stdio.h>
int par[50];
int root(int x){
    if(par[x] == x){return x;}
    return par[x] = root(par[x]);
}

int unite(int x, int y){
    int rx = root(x);
    int ry = root(y);
    if(rx == ry){
        return 0;
    }
    par[rx] = ry;
    return 1;
}

int same(int x, int y){
    int rx = root(x);
    int ry = root(y);
    if(rx == ry){
        return 1;
    }
    return 0;
}
int main(void){
    int N, M;
    int i, j;
    int a[1225], b[1225];
    scanf("%d%d", &N, &M);

    for(i = 0; i < M; i++){
        scanf("%d%d", &a[i], &b[i]);
        a[i]--;
        b[i]--;
    }
    
    int ans = 0;

    for(i = 0; i < M; i++){
        for(j = 0; j < N; j++){
            par[j] = j;
        }
        for(j = 0; j < M; j++){
            if(i == j){
                continue;
            }
            unite(a[j], b[j]);
        }
        if(same(a[i], b[i]) == 0){
            ans++;
        }
    }
    printf("%d\n", ans);
    return 0;   
}