B题是 双联通分量，用Tarjan扫一遍，重新建图然后减掉最深的两条路径 void tarjan(int u, int pre) //Tarjan强连通   {     vis[u] = true;     dfn[u] = low[u] = ++dep;     st.push(u);     fa[u] = true;     for (int i = s[u]; ~i; i = edge[i].nxt)     {         int v = edge[i].v;         if (edge[i].flag == false)              continue;         edge[i].flag = edge[i ^ 1].flag = false;         if (!vis[v])         {             tarjan(v, u);             low[u] = min(low[u], low[v]);             if (dfn[u]<low[v])             {                 bridge[bri_cnt][0] = u;                 bridge[bri_cnt++][1] = v;             }         }         else if (fa[v])              low[u] = min(low[u], dfn[v]);     }     if (dfn[u] == low[u])     {         int t;         do         {             id[t = st.top()] = res;             st.pop();             fa[t] = false;         } while (t != u);         res++;     } } int dfs(int u, int pre) {     int tmp = 0;     for (int i = s[u]; ~i; i = edge[i].nxt)     {         int v = edge[i].v;         if (v == pre) continue;         int d = dfs(v, u);         pre_d = max(pre_d, tmp + d);         tmp = max(tmp, d);     }     return tmp + 1; }