题意:给串s,每次询问k个数a,l个数b,问a和b作为后缀的lcp的综合

题解:和bzoj3879类似,反向sam日神仙...lcp就是fail树上的lca.把点抠出来建虚树,然后在上面dp即可.(感觉之前写的svt什么玩意)

//#pragma GCC optimize(2)

//#pragma GCC optimize(3)

//#pragma GCC optimize(4)

//#pragma GCC optimize("unroll-loops")

//#pragma comment(linker, "/stack:200000000")

//#pragma GCC optimize("Ofast,no-stack-protector")

//#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")

#include<bits/stdc++.h>

#define fi first

#define se second

#define db double

#define mp make_pair

#define pb push_back

#define pi acos(-1.0)

#define ll long long

#define vi vector<int>

#define mod 1000000009

#define ld long double

//#define C 0.5772156649

//#define ls l,m,rt<<1

//#define rs m+1,r,rt<<1|1

#define pll pair<ll,ll>

#define pil pair<int,ll>

#define pli pair<ll,int>

#define pii pair<int,int>

#define ull unsigned long long

//#define base 1000000000000000000

#define fin freopen("a.txt","r",stdin)

#define fout freopen("a.txt","w",stdout)

#define fio ios::sync_with_stdio(false);cin.tie(0)

inline ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}

inline void sub(ll &a,ll b){a-=b;if(a<0)a+=mod;}

inline void add(ll &a,ll b){a+=b;if(a>=mod)a-=mod;}

template<typename T>inline T const& MAX(T const &a,T const &b){return a>b?a:b;}

template<typename T>inline T const& MIN(T const &a,T const &b){return a<b?a:b;}

inline ll qp(ll a,ll b){ll ans=1;while(b){if(b&1)ans=ans*a%mod;a=a*a%mod,b>>=1;}return ans;}

inline ll qp(ll a,ll b,ll c){ll ans=1;while(b){if(b&1)ans=ans*a%c;a=a*a%c,b>>=1;}return ans;}

using namespace std;

const ull ba=233;

const db eps=1e-5;

const ll INF=0x3f3f3f3f3f3f3f3f;

const int N=200000+10,maxn=1000000+10,inf=0x3f3f3f3f;

char s[N];

struct SAM{

    int last,cnt;

    int ch[N<<1][26],fa[N<<1],l[N<<1];

    int pos[N<<1],dep[N<<1],f[N<<1][20],dfn[N<<1],res;

    vi v[N<<1];

    void ins(int c){

        int p=last,np=++cnt;last=np;l[np]=l[p]+1;

        for(;p&&!ch[p][c];p=fa[p])ch[p][c]=np;

        if(!p)fa[np]=1;

        else

        {

            int q=ch[p][c];

            if(l[p]+1==l[q])fa[np]=q;

            else

            {

                int nq=++cnt;l[nq]=l[p]+1;

                memcpy(ch[nq],ch[q],sizeof(ch[q]));

                fa[nq]=fa[q];fa[q]=fa[np]=nq;

                for(;ch[p][c]==q;p=fa[p])ch[p][c]=nq;

            }

        }

    }

    void build()

    {

        last=cnt=1;

        int len=strlen(s+1);

        reverse(s+1,s+1+len);

        for(int i=1;s[i];i++)ins(s[i]-'a'),pos[i]=last;

        for(int i=1;i<=cnt;i++)v[fa[i]].pb(i);

        dfs(1);

        for(int i=1;i<20;i++)for(int j=1;j<=cnt;j++)

            f[j][i]=f[f[j][i-1]][i-1];

    }

    void dfs(int u)

    {

        f[u][0]=fa[u];dfn[u]=++res;

        for(int x:v[u])

        {

//            printf("%d %d\n",u,x);

            dep[x]=dep[u]+1;

            dfs(x);

        }

    }

    int lca(int x,int y)

    {

        if(dep[x]>dep[y])swap(x,y);

        for(int i=19;~i;i--)if(((dep[y]-dep[x])>>i)&1)y=f[y][i];

        if(x==y)return x;

        for(int i=19;~i;i--)if(f[x][i]!=f[y][i])

            x=f[x][i],y=f[y][i];

        return f[x][0];

    }

}sam;

vi v[N*2],in;

void add1(int a,int b){v[a].pb(b),in.pb(a);in.pb(b);}

int st[N*2],top,a[N*2];

ll dp[N*2][2],ans;

void ins(int x)

{

    if(!top){st[++top]=x;return ;}

    int lc=sam.lca(st[top],x);

    while(top>1&&sam.dep[st[top-1]]>sam.dep[lc])

        add1(st[top-1],st[top]),top--;

    if(top>=1&&sam.dep[st[top]]>sam.dep[lc])

        add1(lc,st[top]),top--;

    if(!top||sam.dep[st[top]]<sam.dep[lc])st[++top]=lc;

    st[++top]=x;

}

bool cmp(int a,int b){return sam.dfn[a]<sam.dfn[b];}

void dfs(int u)

{

    ans+=1ll*sam.l[u]*dp[u][0]*dp[u][1];

    for(int x:v[u])

    {

        dfs(x);

        ans+=1ll*sam.l[u]*dp[u][0]*dp[x][1];

        ans+=1ll*sam.l[u]*dp[x][0]*dp[u][1];

        dp[u][0]+=dp[x][0],dp[u][1]+=dp[x][1];

    }

}

int main()

{

    int n,q;scanf("%d%d%s",&n,&q,s+1);

    sam.build();

    while(q--)

    {

        int k,l;scanf("%d%d",&k,&l);

        for(int i=0;i<k;i++)

        {

            scanf("%d",&a[i]),a[i]=sam.pos[n+1-a[i]];

            dp[a[i]][0]++;

        }

        for(int i=0;i<l;i++)

        {

            scanf("%d",&a[k+i]),a[k+i]=sam.pos[n+1-a[k+i]];

            dp[a[k+i]][1]++;

        }

        sort(a,a+k+l,cmp);k+=l;k=unique(a,a+k)-a;

        ans=top=0;ins(1);

        for(int i=0;i<k;i++)if(a[i]!=1)ins(a[i]);

        while(top>=2)add1(st[top-1],st[top]),top--;

        dfs(1);

        printf("%lld\n",ans);

        for(int x:in)v[x].clear(),dp[x][0]=dp[x][1]=0;

        in.clear();

    }

    return 0;

}

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