DFS求无向图环的个数及其最大环的节点数

T组样例，N个结点，M条边（U和V）

无向图的边建两边，cnt记录有多少次走到之前走过的次数(v != fa保证不会往回走)

#include<bits/stdc++.h>
#define IOS ios::sync_with_stdio(false), cin.tie(0),cout.tie(0)
#define endl '\n'
using namespace std;
vector<int> g[100005];
int n, m;
int circle, cnt, max_nodes;
int d[100005];
void dfs(int u, int fa, int deep) {
    d[u] = deep;
    for(auto &v : g[u]) {
        if(v != fa) {
            if(d[v]) {
                cnt ++;
                max_nodes = max(max_nodes, deep - d[v] + 1);
            }
            else {
                dfs(v, u, deep + 1);
            }
        }
    }
}
void solve() {
    cin >> n >> m;
    for(int i = 0; i < m; ++ i) {
        int u, v; cin >> u >> v;
        g[u].push_back(v);
        g[v].push_back(u);
    }
    circle = max_nodes = 0;
    for(int i = 1; i <= n; ++ i) {
        cnt = 0;
        if(!d[i]) {
            dfs(i, -1, 1);
            circle += (cnt == 2);
        }
    }
    cout << circle << ' ' << max_nodes << endl;
    for(int i = 1; i <= n; ++ i) {
        g[i].clear();
        d[i] = 0;
    }
}
signed main(){
    IOS;
    int t = 1;
    cin >> t;
    while(t --) solve();
}