HDU 2222 (AC自动机)

Problem : 给若干个模式串,询问目标串中出现了多少个模式串。
**Solution : **复习了一下AC自动机。需要注意AC自动机中的fail,和next的含义。fail指向了一个最长的与当前匹配出具有相同后缀的一个前缀节点,next用来转移下一个字符,指向最远可以匹配的位置。另外,在目标串匹配时,需要用一个temp指针不断经过fail指针向前跳转,所有经过的节点都是符合的。

#include <iostream>
#include <cstring>
#include <queue>
using namespace std;

const int N = 500008;
struct AC_Automan
{
	int next[N][26];
	int fail[N];
	int cnt[N];
	int root, tot;
	int newnode()
	{
		for (int i = 0; i < 26; ++i) next[tot][i] = -1;
		fail[tot] = -1;
		cnt[tot] = 0;
		return tot++;
	}
	void clear()
	{
		tot = 0;
		root = newnode();
	}
	void insert(const string &s)
	{
		int p = root;
		for (int i = 0, len = s.length(); i < len; ++i)
		{
			if (next[p][s[i] - 'a'] == -1) next[p][s[i] - 'a'] = newnode();
			p = next[p][s[i] - 'a'];
		}
		++cnt[p];
	}
	void build()
	{
		queue <int> Q;
		Q.push(root);
		while (!Q.empty())
		{
			int p = Q.front(); Q.pop();
			for (int i = 0; i < 26; ++i)
			{
				if (~next[p][i])
				{
					if (p == root) fail[next[p][i]] = root;
					else fail[next[p][i]] = next[fail[p]][i];
					Q.push(next[p][i]);
				}
				else
				{	
					if (p == root) next[p][i] = root;
					else next[p][i] = next[fail[p]][i];
				}
			}
		}
	}
	int solve(const string &t)
	{
		int p = root, ans = 0;
		for (int i = 0, len = t.length(); i < len; ++i)
		{
			p = next[p][t[i] - 'a'];
			for (int temp = p; temp != root && cnt[temp] != -1; temp = fail[temp])
			{
				ans += cnt[temp];
				cnt[temp] = -1;
			}
		}
		return ans;
	}
}ac;

int main()
{
	cin.sync_with_stdio(0);
	int T; cin >> T;
	for (int cas = 1; cas <= T; ++cas)
	{
		ac.clear();
		int n; cin >> n;
		for (int i = 0; i < n; ++i)
		{
			string s; cin >> s;
			ac.insert(s);
		}
		ac.build();
		string t; cin >> t;
		cout << ac.solve(t) << endl;
	}
}

原文地址:https://www.cnblogs.com/rpSebastian/p/7191121.html