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 5.4.1) (judge verdict: Accepted)

#include <stdio.h>
#define limit 50

int N,M;
int a[limit],b[limit];

int graph[limit][limit];
int flag[limit];

void dfs(int v){
  flag[v] = 1;
  for(int v2 = 0;v2 < N;v2++){
    if(graph[v][v2] == 0) continue;
    if(flag[v2] == 1) continue;
    dfs(v2);
  }
}

int main(void){
  int ans = 0;
  scanf("%d%d",&N,&M);

  for(int i = 0;i < limit;i++){
    for(int j = 0;j < limit;j++){
      graph[i][j] = 0;
    }
  }
  for(int i = 0;i < M;i++){
    scanf("%d%d",&a[i],&b[i]);
    a[i]--; b[i]--;
    graph[a[i]][b[i]] = graph[b[i]][a[i]] = 1;
  }

  for(int i = 0;i < M;i++){
    graph[a[i]][b[i]] = graph[b[i]][a[i]] = 0;

    for(int j = 0;j < N;j++) flag[j] = 0;

    dfs(0);

    int bridge = 0;
    for(int j = 0;j < N;j++)
      if(flag[j] == 0)
        bridge = 1;
    if(bridge) ans++;

    graph[a[i]][b[i]] = graph[b[i]][a[i]] = 1;
  }

  printf("%d\n",ans);

  return 0;
}