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: Wrong Answer)

#include <stdio.h>
#include <stdbool.h>

int N,M,i,j;
bool graph[55][55];
bool visited[55];
int a[50],b[50];

void dfs(int v)
{
    // 隣接点を探す用
    int v2; 

    // 点vからスタート！
    visited[v] = true;

    for(v2=0;v2<N;v2++){
        // 2点をつなぐ辺が存在するか
        if( graph[v][v2] == false ){
            continue;
        }
        // 訪問済みか
        if( visited[v2] == true ){
            continue;
        }
        // 点vから点v2へ行く辺があり、v2が未訪問　⇒　v2に移動して探索続行
        dfs(v2);
    }

}

int main()
{
    scanf("%d%d",&N,&M);

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

    int count = 0;

    for(i=0;i<M;i++){
        // 辺を一本取り除く
        graph[a[i]][b[i]] = false;
        graph[b[i]][a[i]] = false;

        // すべての点を未訪問に
        for(j=0;j<M;j++){
            visited[j] = false;
        }

        // 深さ優先探索
        dfs(0);

        bool bridge = false;
        for(j=0;j<N;j++){
            if( visited[j] == false ){
                bridge = true;
            }
        }
        if( bridge ){
            count++;
        }

        // 取り除いたのを戻す
        graph[a[i]][b[i]] = true;
        graph[b[i]][a[i]] = true;

    }

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

}