You Are Given Some Strings...【AC自动机】

题目链接：http://codeforces.com/contest/1202/problem/E

枚举文本串的每个位置，计算以该位置为结尾可以匹配多少个模式串，再计算以下一位置为起点能匹配多少模式串，相乘累加作为答案。

#include 
#define rep(i, a, b) for(int i = (a); i <= (b); i++)
#define ll long long
using namespace std;
const int MAX = 750001;
const int N = 500010;
 
const int SIGMA_SIZE = 26;
int num1[MAX],num2[MAX];
struct ACAM {
 
    int ch[MAX][SIGMA_SIZE];
    int val[MAX], last[MAX], f[MAX], sz;
    int ANS;
 
    void init() {
        sz = 1;                         //节点个数
        memset(ch, 0, sizeof(ch));      //路径
        memset(val, 0, sizeof(val));    //结束标记
        memset(f, 0, sizeof(f));        //失配指针
        memset(last, 0, sizeof(last));  //当前节点的上一个模式串
    }
 
    void add(int u) {
        while (u) {
            ANS += val[u];
            u = last[u];
        }
    }
 
    void creat(char *s) {
        int u = 0, len = strlen(s);
        for (int i = 0; i < len; i++) {
            int c = s[i]-'a';
            if (!ch[u][c]) ch[u][c] = sz++;
            u = ch[u][c];
        }
        val[u]++;
    }
 
    void get_fail() {           //找失配指针
        queue q;
        for (int i = 0; i < SIGMA_SIZE; i++)
            if (ch[0][i]) q.push(ch[0][i]);
        while (!q.empty()) {
            int r = q.front();
            q.pop();
            for (int c = 0; c < SIGMA_SIZE; c++) {
                int u = ch[r][c];
                if (!u) {
                    ch[r][c] = ch[f[r]][c];
                    continue;
                }
                q.push(u);
                int v = f[r];
                while (v && ch[v][c] == 0) v = f[v];
                f[u] = ch[v][c];
                last[u] = val[f[u]] ? f[u] : last[f[u]];
            }
        }
    }
 
    void find(char *T,int oper) {
        int len = strlen(T), j = 0;
        for (int i = 0; i < len; i++) {
            ANS = 0;
            int c = T[i]-'a';
            j = ch[j][c];
            add(j);
            if(oper==1) num1[i] = ANS;
            else num2[i] = ANS;
        }
    }
}AC1,AC2;
char s[N],k[N];
int main() {
    //freopen("a.txt","r",stdin);
    ios::sync_with_stdio(false);
    cin>>s;
    int n;
    cin>>n;
    AC1.init();
    AC2.init();
    rep(i, 1, n) {
        cin>>k;
        AC1.creat(k);
        int len = strlen(k);
        reverse(k+0,k+len);
        AC2.creat(k);
    }
    AC1.get_fail();
    AC2.get_fail();
    AC1.find(s,1);
 
    int len = strlen(s);
    reverse(s+0,s+len);
    AC2.find(s,2);
 
    ll ans = 0;
    rep(i, 0, len-2)
        ans +=  1ll*num1[i]*num2[len-2-i];
    cout<