E. You Are Given Some Strings...

E. You Are Given Some Strings...

AC自动机

求一个串$t$中包含子串$s_{i}+s_{j}$的个数。

可以正反跑两遍AC自动机

正着跑，表示$s_{i}$结束，反正跑对应$s_{i}$开头

#include<bits/stdc++.h>
using namespace std;
#define maxn 1000005
#define maxm 28
struct AC{
int trieN;
int ch[maxn][maxm];
int val[maxn],tt[maxn];
int fail[maxn];
int sum[maxn];
vector<int> v;
void init()
{
    trieN=-;
    newnod();
}
int newnod()
{
    memset(ch[++trieN],,sizeof ch[]);
    val[trieN]=fail[trieN]=;
    return trieN;
}
void insert(const string & str)
{
    int cur=;
    for(int i=;i<str.size();i++){
        int d=str[i]-'a';
        if(!ch[cur][d]){
            ch[cur][d]=newnod();
        }
        cur=ch[cur][d];
    }
    val[cur]++;
}
void build()
{
    queue<int> q;
    for(int i=;i<maxm;i++){
        if(ch[][i]){
            q.push(ch[][i]);
        }
    }
    while(!q.empty()){
        int cur=q.front();
        v.push_back(cur);
        q.pop();
        for(int i=;i<maxm;i++){
            if(ch[cur][i]){
                fail[ch[cur][i]]=ch[fail[cur]][i];
                q.push(ch[cur][i]);
            }else{
                ch[cur][i]=ch[fail[cur]][i];
            }
        }
    }
    for(int i=;i<v.size();i++){
        int u=v[i];
        tt[u]=tt[fail[u]]+val[u];
    }///优化？？
}

void  query(const string & str)
{

    int res=,cur=;
    for(int i=;str[i];i++){
        int d=str[i]-'a';
        cur=ch[cur][d];

        sum[i]=tt[cur];
        res=;
    }
    //return res;
}
}ac1,ac2;

int main()
{
    string s,t;
    cin>>t;
    int n;
    scanf("%d",&n);
    for(int i=;i<n;i++){
        cin>>s;

        ac1.insert(s);
        reverse(s.begin(),s.end());
        ac2.insert(s);
    }
    ac1.build();
    ac2.build();
    ac1.query(t);
    long long ans=;
    reverse(t.begin(),t.end());
    ac2.query(t);
    int len=t.length();
    /*for(int i=0;i<len;i++){
        cout<<ac1.sum[i]<<" "<<ac2.sum[i]<<endl;
    }*/
   // cout<<len<<'\n';
    for(int i=;i<=len;i++){
        ans+=(long long)ac1.sum[i-]*ac2.sum[len-i-];
    }
    cout<<ans<<'\n';
}