【题解】SHOI2008仙人掌图

　　本质上还是树形dp。建立圆方树，遇到圆点的时候直接求（和树形dp一样即可），遇到方点做中转点的时候要考虑会从圆的另一侧通过（需满足最短路径的原则）。原本是对于圆上的点进行 \(n^{2}\) 的匹配，果断超时。但没有发现 \(n ^ {2}\) 的dp明显是一个可以单调队列优化的dp。所以在遇上难解决的问题的时候，一定要融会贯通地思考。有一个细节：将圆复制一下可以去掉 \(max\)造成的影响，dp就十分方便了。

#include <bits/stdc++.h>
using namespace std;
#define maxn 500000
int n, m, N, timer, ans;
int fa[maxn], dfn[maxn], low[maxn];
int f[maxn][], id[maxn], S[maxn];
int tmp[maxn], q[maxn];
int F[maxn][], SS[maxn];
 
struct edge
{
    int cnp, head[maxn], to[maxn], last[maxn], w[maxn];
    edge() { cnp = ; }
    void add(int u, int v, int ww)
    {
        to[cnp] = v, last[cnp] = head[u], w[cnp] = ww, head[u] = cnp ++;
        to[cnp] = u, last[cnp] = head[v], w[cnp] = ww, head[v] = cnp ++;
    }
}E1, E2;
 
int read()
{
    int x = , k = ;
    char c;
    c = getchar();
    while(c < '' || c > '') { if(c == '-') k = -; c = getchar(); }
    while(c >= '' && c <= '') x = x *  + c - '', c = getchar();
    return x * k;
}
 
void Solve(int u, int v)
{
    ++ N; int ID = , pre = ;
    for(int i = v; i != fa[u]; i = fa[i])
        S[i] = pre ++, id[i] = ++ ID;
    S[N] = S[u], S[u] = ;
    for(int i = v; i != fa[u]; i = fa[i])
        E2.add(N, i, min(S[i], S[N] - S[i]));
}
 
void Tarjan(int u)
{
    dfn[u] = low[u] = ++ timer;
    for(int i = E1.head[u]; i; i = E1.last[i])
    {
        int v = E1.to[i];
        if(v == fa[u]) continue;
        if(!dfn[v])
        {
            fa[v] = u; Tarjan(v);
            low[u] = min(low[u], low[v]);
        }
        else low[u] = min(low[u], dfn[v]);
        if(low[v] > dfn[u]) E2.add(u, v, );
    }
    for(int i = E1.head[u]; i; i = E1.last[i])
    {
        int v = E1.to[i];
        if(dfn[u] < dfn[v] && fa[v] != u) Solve(u, v);
    }
}
 
void update(int u, int w)
{
    if(w <= f[u][]) return;
    if(w < f[u][]) f[u][] = w;
    else f[u][] = f[u][], f[u][] = w;
}
 
bool cmp(int a, int b) { return id[a] < id[b]; }
 
int Work(int u)
{
    int head = , tail = , cnt = , ans = , len = ;
    for(int i = E2.head[u]; i; i = E2.last[i])
        tmp[++ cnt] = f[E2.to[i]][], len ++;
    for(int i = ; i <= len; ++ i)
        tmp[++ cnt] = tmp[i];
    q[head] = ;
    for(int i = ; i <= cnt; ++ i)
    {
        while(head < tail && i - q[head] > len / ) ++ head;
        ans = max(ans, tmp[i] + tmp[q[head]] + i - q[head]);
        while(head <= tail && tmp[i] - i > tmp[q[tail]] - q[tail] ) -- tail;
        q[++ tail] = i;
    }
    return ans;
}
 
void dfs(int u, int ff)
{
    for(int i = E2.head[u]; i; i = E2.last[i])
    {
        int v = E2.to[i];
        if(v == ff) continue; else dfs(v, u);
        update(u, f[v][] + E2.w[i]);
    }
    if(u <= n) ans = max(ans, f[u][] + f[u][]);
    else ans = max(ans, Work(u));
}
 
int main()
{
    N = n = read(), m = read();
    for(int i = ; i <= m; i ++)
    {
        int tot = read(), u, pre;
        for(int j = ; j <= tot; j ++)
        {
            u = read(); if(j == ) { pre = u; continue; }
            E1.add(u, pre, ); pre = u;
        }
    }
    Tarjan();
    dfs(, );
    printf("%d\n", ans);
    return ;
}