一个AC自动机模版

先输入一堆子串，再输入一个母串

询问有多少个子串在母串中出现过

（模仿的洛谷大佬的板子

//
// Created by xingchaoyue on 2019/4/19.
//

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<queue>
const int maxn = 5000000+10;
using namespace std;
char str[maxn*2];
struct node{
    int fail;
    int cnt;
    int next[27];
}tree[maxn];

int k=0,ans=0;
queue<int>q;
void build_tree(int id,char *s){
    int len = strlen(s);
    int j=0;
    for(int i =0;i<len;++i){
        j = s[i]-'a';
        if(tree[id].next[j]==0){
            tree[id].next[j]=++k;
        }
        id = tree[id].next[j];
    }
    tree[id].cnt++;
}

void build_fail(int id){
    while(!q.empty())q.pop();
    for(int i =0;i<26;++i){
        int j = tree[id].next[i];
        if(j!=0){
            q.push(j);
            tree[j].fail = id;
        }
    }
    while(!q.empty()){
        int now = q.front();
        q.pop();
        for(int i =0;i<26;++i){
            int j = tree[now].next[i];
            if(j==0){
                tree[now].next[i] = tree[tree[now].fail].next[i];
                continue;

            }

                tree[j].fail = tree[tree[now].fail].next[i];
                q.push(j);

        }
    }
}

void solve(int id,char *s){
    int len = strlen(s),j = 0;
    for(int i =0;i<len;++i){
        int j = tree[id].next[s[i]-'a'];
        while(j&&tree[j].cnt!=-1){
            ans+=tree[j].cnt;
            tree[j].cnt=-1;
            j = tree[j].fail;
        }
        id = tree[id].next[s[i]-'a'];
    }
}

int main(){
    int n;
    scanf("%d",&n);
    for(int i =1;i<=n;++i){
        scanf("%s",str);
        build_tree(0,str);
    }
    build_fail(0);
    scanf("%s",str);
    solve(0,str);
    printf("%d\n",ans);
    return 0;
}