[SHOI2008]仙人掌图

题目链接
考虑用\(tarjan\)找环,环内\(dp\),环外\(dp\).
用\(f[u]\)表示到\(u\)点的最长距离长度.
如果我们找到一条边是桥就直接转移——\(f[u]=max(f[u],f[v]+len)\),同时更新\(Ans\).
我们其实要求的就是\(max(f[i]+f[j]+dis(i,j))\)
如果我们找到一个环,首先用环中节点的\(f\)值去更新\(Ans\).
首先破环成链,然后这显然是一个区间\(dp\).
我们用单调队列去优化这个\(dp\)
如果队首和现在节点的距离大于\(\frac{Len_{cir}}{2}\),就\(pop\).

然后把这个环的答案都并到环的起始点即可.
代码如下

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<cmath>
#include<vector>
#define N (1000010)
#define M (20000010)
#define inf (0x7f7f7f7f)
#define rg register int
#define Label puts("NAIVE")
#define spa print(' ')
#define ent print('\n')
#define rand() (((rand())<<(15))^(rand()))
typedef long double ld;
typedef long long LL;
typedef unsigned long long ull;
using namespace std;
inline char read(){
    static const int IN_LEN=1000000;
    static char buf[IN_LEN],*s,*t;
    return (s==t?t=(s=buf)+fread(buf,1,IN_LEN,stdin),(s==t?-1:*s++):*s++);
}
template<class T>
inline void read(T &x){
    static bool iosig;
    static char c;
    for(iosig=false,c=read();!isdigit(c);c=read()){
        if(c=='-')iosig=true;
        if(c==-1)return;
    }
    for(x=0;isdigit(c);c=read())x=((x+(x<<2))<<1)+(c^'0');
    if(iosig)x=-x;
}
inline char readchar(){
    static char c;
    for(c=read();!isalpha(c);c=read())
    if(c==-1)return 0;
    return c;
}
const int OUT_LEN = 10000000;
char obuf[OUT_LEN],*ooh=obuf;
inline void print(char c) {
    if(ooh==obuf+OUT_LEN)fwrite(obuf,1,OUT_LEN,stdout),ooh=obuf;
    *ooh++=c;
}
template<class T>
inline void print(T x){
    static int buf[30],cnt;
    if(x==0)print('0');
    else{
        if(x<0)print('-'),x=-x;
        for(cnt=0;x;x/=10)buf[++cnt]=x%10+48;
        while(cnt)print((char)buf[cnt--]);
    }
}
inline void flush(){fwrite(obuf,1,ooh-obuf,stdout);}
int n,m,f[N],fi[N],ne[M],b[M],fa[N],cir[N];
int h,t,q[N*2],ans,dep[N],dfn[N],low[N],ind,E;
void dp(int u,int st){
    int len=0,h=1,t=0;
    for(int i=u;i!=st;i=fa[i])cir[++len]=i;cir[++len]=st;
    for(int i=1;i<=len/2;i++)swap(cir[i],cir[len-i+1]);
    for(int i=1;i<=len;i++)cir[i+len]=cir[i];
    for(int i=1;i<=len*2;i++){
        while(h<=t&&i-q[h]>len/2)h++;
        if(h<=t)ans=max(ans,f[cir[q[h]]]+f[cir[i]]+i-q[h]);
        while(h<=t&&f[cir[q[t]]]<f[cir[i]])t--;
        q[++t]=i;
    }
    for(int i=u;i!=st;i=fa[i]){
        f[st]=max(f[st],f[i]+min(dep[i]-dep[st],dep[u]-dep[i]+1));
    }
}
void tarjan(int u,int pre){
    dfn[u]=low[u]=++ind,fa[u]=pre,dep[u]=dep[pre]+1;
    for(int i=fi[u];i;i=ne[i]){
        int v=b[i];
        if(!dfn[v]){
            tarjan(v,u);
            low[u]=min(low[u],low[v]);
            if(low[v]>dfn[u])
            ans=max(ans,f[u]+f[v]+1),f[u]=max(f[u],f[v]+1);
        }
        else if(v!=pre)low[u]=min(low[u],dfn[v]);
    }
    for(int i=fi[u];i;i=ne[i]){
        int v=b[i];
        if(fa[v]!=u&&dfn[v]>dfn[u])
        dp(v,u);
    }
}
void add(int x,int y){
    ne[++E]=fi[x],fi[x]=E,b[E]=y;
}
int main(){
    read(n),read(m);
    for(int i=1;i<=m;i++){
        int k,s; read(k),read(s);
        for(int j=1;j<k;j++){
            int x; read(x);
            add(s,x),add(x,s),s=x;
        }
    }
    tarjan(1,0);
    for(int i=1;i<=n;i++)ans=max(ans,f[i]);
    printf("%d\n",ans);
}