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>
#include <string.h>
#include <math.h>

int descending_compare(const void *a, const void *b){
    if (*(int*)a > *(int*)b){
        return -1;
    }else if (*(int*)a == *(int*)b){
        return 0;
    }else{
        return 1;
    }
}

int ascending_compare(const void *a, const void *b){
    if (*(int*)a < *(int*)b){
        return -1;
    }else if (*(int*)a == *(int*)b){
        return 0;
    }else{
        return 1;
    }
}


int lower_bound(int *a, int n, int key){
    int left, mid, right;
    left = 0, right = n;
    mid = (left + right)/2;
    while ((left+1 != mid || mid+1 != right) && mid != left){
        if (key > a[mid]){
            left = mid;
        }else{
            right = mid+1;
        }
        mid = (left + right)/2;
    }
    if (a[left] >= key)return left;
    if (a[mid] >= key)return mid;
    if (a[right] >= key)return right;
    return n;
}

int queue[1005];
int v[51][51] = {0};
int node[51];

int main(void){
    int head, tail;
    int n, m;
    int a, b;
    int i, j, k;
    int ans = 0;
    int flag;
    scanf("%d %d", &n, &m);
    for (i = 0; i < m; i++){
        scanf("%d %d", &a, &b);
        v[a][b] = 1;
        v[b][a] = 1;
    }

    for (i = 0; i < 51*51; i++){
        if (v[i/51][i%51] == 1){
            flag = 1;
            for (j = 0; j < 51; j++){
                node[j] = 0;
            }
            v[i/51][i%51] = 0;
            head = 0;
            tail = 0;
            queue[tail] = (i/51);
            tail++;
            while (head < tail){
                for (k = 0; k < 51; k++){
                    if (v[queue[head]][k] == 1){
                        if (node[k] != 1){
                            queue[tail] = k;
                            node[k] = 1;
                            tail++;
                            if (k == i%51){
                                flag = 0;
                            }
                        }
                    }
                }
                head++;
            }
            if (flag == 1){
                ans++;
            }
            v[i/51][i%51] = 1;
        }
    }
    printf("%d\n", ans/2);
    return 0;    
    
}