AC自动机（二次加强版）

题目链接 P5357

  要找每个串的出现次数，实际上就是在fail树上进行处理了，我们知道，在fail树的祖先节点上的点，一定是在这之前的前缀的点，所以直接进行跳转就可以了。

#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 = 2e5 + 7;
int N, tot, root;
struct Trie_node
{
    int nex[26], fail; vector<int> val;
    Trie_node() { memset(nex, 0, sizeof(nex)); val.clear(); fail = 0; }
    void clear() { memset(nex, 0, sizeof(nex)); val.clear(); fail = 0; }
} t[maxN];
char s[maxN], T[2000006];
void Insert(int ith)
{
    int len = (int)strlen(s), u = root;
    for(int i=0, id; i<len; i++)
    {
        id = s[i] - 'a';
        if(!t[u].nex[id])
        {
            t[u].nex[id] = ++tot;
            t[tot].clear();
        }
        u = t[u].nex[id];
    }
    t[u].val.push_back(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];
                }
                que[tail++] = son;
            }
            else t[tmp].nex[i] = t[t[tmp].fail].nex[i];
        }
    }
}
vector<int> to[maxN];
int siz[maxN] = {0}, maxx;
int ans[maxN];
void dfs(int u)
{
    for(int v : to[u])
    {
        dfs(v);
        siz[u] += siz[v];
    }
    if(!t[u].val.empty())
    {
        for(int i : t[u].val)
        {
            ans[i] = siz[u];
        }
    }
}
int main()
{
    scanf("%d", &N);
    tot = 0; root = 0;
    t[root].clear();
    for(int i=1; i<=N; i++)
    {
        scanf("%s", s);
        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;
    dfs(root);
    for(int i=1; i<=N; i++) printf("%d
", ans[i]);
    return 0;
}