【模板】AC自动机（加强版）【AC自动机fail树上求最多出现次数】

题目链接 P3796

  给出N个模式串，然后我们用一个文本串去进行匹配，这样的做法，就是AC自动机了，于是乎，我们可以先将N个模式串丢进去，然后建立fail树，然后先对所有的节点求出最大串在文本串中出现的次数，然后利用dfs跑fail树的办法，我们可以O(N)的求解这个问题。

#include <iostream>
#include <cstdio>
#include <cmath>
#include <string>
#include <cstring>
#include <algorithm>
#include <limits>
#include <vector>
#include <stack>
#include <queue>
#include <set>
#include <map>
#include <bitset>
#include <unordered_map>
#include <unordered_set>
#define lowbit(x) ( x&(-x) )
#define pi 3.141592653589793
#define e 2.718281828459045
#define INF 0x3f3f3f3f
#define HalF (l + r)>>1
#define lsn rt<<1
#define rsn rt<<1|1
#define Lson lsn, l, mid
#define Rson rsn, mid+1, r
#define QL Lson, ql, qr
#define QR Rson, ql, qr
#define myself rt, l, r
#define pii pair<int, int>
#define MP(a, b) make_pair(a, b)
using namespace std;
typedef unsigned long long ull;
typedef unsigned int uit;
typedef long long ll;
const int maxN = 70 * 150 + 7;
int N, tot, root;
struct Trie_node
{
    int nex[26], val, fail;
    Trie_node() { memset(nex, 0, sizeof(nex)); val = 0; fail = 0; }
    void clear() { memset(nex, 0, sizeof(nex)); val = 0; fail = 0; }
} t[maxN];
char s[155][77], T[1000006];
void Insert(int ith)
{
    int len = (int)strlen(s[ith]), u = root;
    for(int i=0, id; i<len; i++)
    {
        id = s[ith][i] - 'a';
        if(!t[u].nex[id])
        {
            t[u].nex[id] = ++tot;
            t[tot].clear();
        }
        u = t[u].nex[id];
    }
    t[u].val = ith;
}
int que[maxN], top, tail;
void build_fail()
{
    top = tail = 0;
    que[tail++] = root;
    int tmp, p, son;
    while(top < tail)
    {
        tmp = que[top++];
        for(int i=0; i<26; i++)
        {
            son = t[tmp].nex[i];
            if(son)
            {
                if(!tmp) t[son].fail = 0;
                else
                {
                    p = t[tmp].fail;
                    while(p && !t[p].nex[i]) p = t[p].fail;
                    t[son].fail = t[p].nex[i];
                }
                if(t[t[son].fail].val) t[son].val += t[t[son].fail].val;
                que[tail++] = son;
            }
            else t[tmp].nex[i] = t[t[tmp].fail].nex[i];
        }
    }
}
vector<int> to[maxN];
int siz[maxN] = {0}, maxx;
vector<int> ans;
void dfs(int u)
{
    for(int v : to[u])
    {
        dfs(v);
        siz[u] += siz[v];
    }
    if(t[u].val)
    {
        if(siz[u] > maxx)
        {
            maxx = siz[u];
            ans.clear();
            ans.push_back(t[u].val);
        }
        else if(siz[u] == maxx) ans.push_back(t[u].val);
    }
}
int main()
{
    while(scanf("%d", &N) && N)
    {
        tot = 0; root = 0;
        t[root].clear();
        for(int i=1; i<=N; i++)
        {
            scanf("%s", s[i]);
            Insert(i);
        }
        build_fail();
        for(int i=0; i<=tot; i++) { to[i].clear(); siz[i] = 0; }
        for(int i=1; i<=tot; i++) to[t[i].fail].push_back(i);
        scanf("%s", T);
        int u = root, len = (int)strlen(T);
        for(int i=0, id; i<len; i++)
        {
            id = T[i] - 'a';
            while(u && !t[u].nex[id]) u = t[u].fail;
            u = t[u].nex[id];
            siz[u]++;
        }
        maxx = 0; ans.clear();
        dfs(root);
        printf("%d\n", maxx);
        sort(ans.begin(), ans.end());
        for(int i : ans)
        {
            printf("%s\n", s[i]);
        }
    }
    return 0;
}