#include <bits/stdc++.h> // By Lucky Ox
#define endl "\n"
#define int long long
#define PI atan(1.0) * 4
#define pii pair<int, int>
using namespace std;
using i128 = __int128;
using ull = unsigned long long;
const int mod = LONG_LONG_MAX, INF = 0x3f3f3f3f3f3f3f3f;
int P(int x, int p){ return (x % p + p) % p; }
int lcm(int x, int y) { return x / gcd(x, y) * y; }
int gcd(int a, int b) { return b == 0 ? a : gcd(b, a % b); }
int len_10(int x) { int len = 0; while(x) { x /= 10; len ++ ; } return len; }
int q_pow(int a, int k, int p) { int res = 1; while (k) { if (k & 1) res = res * a % p; k >>= 1; a = a * a % p; } return res; }
int to_int(string s) { int val = 0; for(int i = 0; i < (int)s.size(); i ++ ){val *= 10; val += s[i] - '0';}return val;}//注意:s是空串也会返回0
i128 read() { i128 x = 0; char c = getchar(); while (c < '0' || c > '9') c = getchar(); while (c >= '0' && c <= '9') {
x = x * 10 + c - '0'; c = getchar(); } return x; } //i128输入
void print(i128 x) { if(x > 9) print(x / 10); putchar(x % 10 + '0'); }//i128输出
//用__lg()来求一个数二进制下的位数 返回的len 表示这个数是[0, 1, ...., len] 比如10 __lg(10) = 3, 1010 [3, 2, 1, 0]
//__builtin_popcountll(int x) 求二进制下x中1的数量 __buitlin_ctzll(int x) 求二进制下末尾0的个数
//将x转换成[1, n]中对应的数 (x - 1) % n + 1

const int N = 5e3 + 10;
int n, m;
vector<int> e[N];
int d[N], low[N], idx;

void dfs(int u, int fa) {
    low[u] = ++ idx;
    for(int v : e[u]) {
        if(v == fa) continue;

        if(!low[v]) dfs(v, u);
        low[u] = min(low[u], low[v]);
    }
}

int tarjan() {
    for(int u = 1; u <= n; u ++ ) 
        for(int v : e[u]) 
            if(low[u] != low[v]) d[low[u]] ++ ;

    int res = 0;
    for(int i = 1; i <= n; i ++ ) 
        if(d[i] == 1) res ++ ;
    return res;
}

void solve() {
    cin >> n >> m;
    for(int i = 1; i <= m; i ++ ) {
        int u, v;
        cin >> u >> v;
        e[u].push_back(v); e[v].push_back(u);
    }

    for(int i = 1; i <= n; i ++ ) {
        if(!low[i]) dfs(i, 0);
    }

    int ans = tarjan();
    cout << (ans + 1) / 2 << endl;
}

signed main() {
    ios::sync_with_stdio(0);cin.tie(0);
    cout << fixed << setprecision(10);
    int T = 1;
    while (T -- ) solve();
    return 0;
}

代码如上 很少写有重边的问题 突然想不到了