poj 3987 Computer Virus on Planet Pandora ac自动机复习

• 题意如下
  给出多个模式串，最后给出一个文本串，求有多少个模式串被文本串包含或者被反序的文本串包含

• 几乎是ac自动机模板，正反跑两次就行了。但是当时板子怎么写忘记了，队里也没带资料，于是没调出来。说明还是基本功不够熟练

代码如下

#include <cstdio>
#include <iostream>
#include <algorithm>
#include <cstring>
using namespace std;


int t, n, co = 0, ans = 0;

char pat[1005];
char ss[5100005];
char ss2[5100005];
int que[500005];
bool vis[500005];
struct trie
{
	int fl;
	int fail;
	int nxt[26];
} aca[500005];

void insert(char *s, int l)
{
	int o = 0;
	int p = 0;
	int ls = strlen(s);
	while (p < ls)
	{
		if (!aca[o].nxt[s[p] - 'A'])
		{
			aca[o].nxt[s[p] - 'A'] = ++co;
		}
		o = aca[o].nxt[s[p] - 'A'];
		++p;
	}
	++aca[o].fl;
}

void fail_p()
{
	int hd = 1, tl = 1;
	for (int i = 0; i < 26; ++i)
	{
		if (aca[0].nxt[i])
		{
			que[tl++] = aca[0].nxt[i];
			aca[aca[0].nxt[i]].fail = 0;
		}
	}
	while (hd != tl)
	{
		int u = que[hd++];
		for (int i = 0; i < 26; ++i)
		{
			if (aca[u].nxt[i])
			{
				aca[aca[u].nxt[i]].fail = aca[aca[u].fail].nxt[i];
				que[tl++] = aca[u].nxt[i];
			}
			else
			{
				aca[u].nxt[i] = aca[aca[u].fail].nxt[i];
			}
		}
	}
}

void query(char *s)
{
	int o = 0;
	int l = strlen(s);
	vis[0] = true;
	for (int i = 0; i < l; ++i)
	{
		int sn = aca[o].nxt[s[i] - 'A'];
		while (!vis[sn])
		{
			ans += aca[sn].fl;
			vis[sn] = true;
			sn = aca[sn].fail;
		}
		o = aca[o].nxt[s[i] - 'A'];
	}
	o = 0;
	for (int i = l - 1; i >= 0; --i)
	{
		int sn = aca[o].nxt[s[i] - 'A'];
		while (sn && !vis[sn])
		{
			ans += aca[sn].fl;
			vis[sn] = true;
			sn = aca[sn].fail;
		}
		o = aca[o].nxt[s[i] - 'A'];
	}
}

int main()
{
	scanf("%d", &t);
	while (t--)
	{
		aca[0].fl = aca[0].fail = 0;
		for (int i = 0; i < 26; ++i)
		{
			aca[0].nxt[i] = 0;
		}
		co = 0;
		ans = 0;
		memset(vis, 0, sizeof(vis));
		memset(aca, 0, sizeof(aca));
		scanf("%d", &n);
		for (int i = 1; i <= n; ++i)
		{
			scanf("%s", pat);
			insert(pat, i);
		}
		fail_p();
		scanf("%s", &ss2);
		int ll = -1;
		int l2 = strlen(ss2);
		for (int i = 0; i < l2; ++i)
		{
			if (ss2[i] != '[')
			{
				ss[++ll] = ss2[i];
			}
			else
			{
				int x = 0;
				int j;
				for (j = 1; ss2[i + j + 1] != ']'; ++j)
				{
					x = x * 10 + ss2[i + j] - '0';
				}
				char c = ss2[i + j];
				i += j + 1;
				while (x)
				{
					ss[++ll] = c;
					--x;
				}
			}
		}
		ss[++ll] = 0;
		query(ss);
		printf("%d\n", ans);
	}
	return 0;
}