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;
}