思路：先求出非环上的最长链，然后环上用单调队列优化dp。

代码：

#pragma GCC optimize(2)#pragma GCC optimize(3)#pragma GCC optimize(4)#include
    
     using namespace std;#define y1 y11#define fi first#define se second#define pi acos(-1.0)#define LL long long//#define mp make_pair#define pb push_back#define ls rt<<1, l, m#define rs rt<<1|1, m+1, r#define ULL unsigned LL#define pll pair
     
      #define pli pair
      
       #define pii pair
       
        #define piii pair
        
         #define pdd pair
         
          #define mem(a, b) memset(a, b, sizeof(a))#define debug(x) cerr << #x << " = " << x << "\n";#define fio ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);//headconst int N = 5e4 + 5;vector
          
            g[N];int fa[N], deep[N], dp[N], a[N*2], n, m, t, ans;int dfn[N], low[N], cnt;inline void DP(int u, int v) { int t = 0; for (int i = v; i != u; i = fa[i]) a[++t] = i; a[++t] = u; for (int i = 1; i <= t; ++i) a[i+t] = a[i]; deque
           
             q; for (int i = 1; i <= 2*t; ++i) { while(!q.empty() && i-q.front() > t/2) q.pop_front(); if(!q.empty()) ans = max(ans, dp[a[i]]+dp[a[q.front()]]+i-q.front()); while(!q.empty() && dp[a[q.back()]]-q.back() <= dp[a[i]]-i) q.pop_back(); q.push_back(i); } for (int i = v; i != u; i = fa[i]) { dp[u] = max(dp[u], dp[i]+min(deep[i]-deep[u], deep[v]-deep[i]+1)); }}inline void tarjan(int u, int o) { fa[u] = o; deep[u] = deep[o]+1; dfn[u] = low[u] = ++cnt; for (int i = 0; i < g[u].size(); ++i) { int v = g[u][i]; if(v == o) continue; if(!dfn[v]) { tarjan(v, u); low[u] = min(low[u], low[v]); } else low[u] = min(low[u], dfn[v]); if(low[v] > dfn[u]) { ans = max(ans, dp[u]+dp[v]+1); dp[u] = max(dp[u], dp[v]+1); } } for (int i = 0; i < g[u].size(); ++i) { int v = g[u][i]; if(fa[v] != u && dfn[v] > dfn[u]) { DP(u, v); } }}int main() { while(~scanf("%d %d", &n, &m)) { cnt = ans = 0; for (int i = 1; i <= m; ++i) { scanf("%d", &t); for (int i = 1; i <= t; ++i) scanf("%d", &a[i]); for (int i = 2; i <= t; ++i) g[a[i-1]].pb(a[i]), g[a[i]].pb(a[i-1]); } tarjan(1, 0); printf("%d\n", ans); for (int i = 1; i <= n; ++i) g[i].clear(), low[i] = dfn[i] = dp[i] = 0; } return 0;}