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++14 (GCC 5.4.1) (judge verdict: Accepted)


#include <map>
#include <set>
#include <cmath>
#include <queue>
#include <stack>
#include <cstdio>
#include <string>
#include <vector>
#include <cstdlib>
#include <cstring>
#include <sstream>
#include <iostream>
#include <algorithm>
#include <functional>
#include <iostream>



using namespace std;

int p[51];

int fxr[51];

int finds(int x)
{
        if(p[x]==x)return x;
        else return p[x] = finds(p[x]);
}

bool same(int x, int y)
{
        return (finds(x) == finds(y));
}

void unions (int x,int y)
{
        x = finds(x);
        y = finds(y);
        if(x == y)return;
        else if(fxr[x] > fxr[y])
        {
                p[y] = x;
        }
        else
        {
                if(fxr[x] == fxr[y])
                fxr[y]++;
                p[x] = y;
        }
}

int main(){
        int N,M,cnt,ans=0;
        cin >> N >> M;
        int t[M],k[M];
        for(int i=0;i<M;i++)
        {
                cin >> t[i] >> k[i];
        }
        for(int i=0;i<M;i++)
        {
                cnt = 0;
                for(int j=0;j<=N;j++)
                {
                        p[j] = j;
                        fxr[j] = 0;
                }
                for(int j=0;j<M;j++)
                {
                        if(i != j)
                        unions(t[j],k[j]);
                }
                cnt = 0;
                for(int j=1;j<=N;j++)
                {
                if(same(1,j) == true)
                cnt++;
                }
                if(cnt!=N)
                ans++;
        }
        cout << ans << endl;
        return 0;
}