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>
#include<stdlib.h>

typedef long long ll;

int max(int a, int b){
  if(a > b) return a;
  else return b;
}
int min(int a, int b){
  if(a < b) return a;
  else return b;
}

  int seen[51] = {0};
  int n,m;
  int graph[51][51] = {0};
  int flag = 0;
  int count = 0;

void dfs(int x, int y){
  if(seen[y] == 1) return;

  seen[y] = 1;

  for(int i = 1; i <= n; ++i){
    if(seen[i] == 0 && graph[y][i] == 1){
      dfs(y,i);
    }
  }
}

int main(){


  scanf("%d %d",&n,&m);

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

  int flag2 = 0;
  int x,y;
  for(int i = 1; i <= n; ++i){
    for(int j = i+1; j <= n; ++j){
      if(graph[i][j] == 1 || graph[j][i] == 1){
        if(flag2 == 0){
          x = i;
          y = j;
          flag2 = 1;
        }
        graph[i][j] = 0;
        graph[j][i] = 0;
          dfs(x,y);
          for(int k = 1; k <= n; ++k){
            if(seen[k] == 0) flag = 1;
            seen[k] = 0;
          }
          if(flag == 1) count ++;
        graph[i][j] = 1;
        graph[j][i] = 1;
        flag = 0;
      }
    }
  }

  printf("%d",count);

  return 0;

}