HDU 4416 (后缀自动机)

HDU 4416 Good Article Good sentence

Problem : 给一个串S，和一些串T，询问S中有多少个子串没有在T中出现。
Solution :首先对所有的T串建立后缀自动机，统计出本质不同的子串个数ans1，再将S串插入后缀自动机，统计出本质不同的子串个数ans2，则答案即为ans2-ans1。
将多个串插入后缀自动机，只需要每次将last赋值为root即可。

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

const int N = 1000008;

struct Suffix_Automaton
{
	int nt[N << 1][26], a[N << 1], fail[N << 1];
	int tot, last, root;
	int p, q, np, nq;
	int newnode(int len)
	{
		for (int i = 0; i < 26; ++i) nt[tot][i] = -1;
		fail[tot] = -1; a[tot] = len; 
		return tot++;
	}
	void clear()
	{
		tot = last = 0; 
		root = newnode(0);
	}
	void insert(int ch)
	{
		p = last; last = np = newnode(a[p] + 1); 
		for (; ~p && nt[p][ch] == -1; p = fail[p]) nt[p][ch] = np;
		if (p == -1) fail[np] = root;
		else
		{
			q = nt[p][ch];
			if (a[p] + 1 == a[q]) fail[np] = q;
			else
			{
				nq = newnode(a[p] + 1);
				for (int i = 0; i < 26; ++i) nt[nq][i] = nt[q][i];
				fail[nq] = fail[q];
				fail[np] = fail[q] = nq;
				for (; ~p && nt[p][ch] == q; p = fail[p]) nt[p][ch] = nq;
			}
		}
	}
	long long solve()
	{
		long long ans = 0;
		for (int i = 1; i < tot; ++i) ans += a[i] - a[fail[i]];
		return ans;
	}
}sam;


int main()
{
	cin.sync_with_stdio(0);
	int T; cin >> T;
	for (int cas = 1; cas <= T; ++cas)
	{
		sam.clear();
		int n; cin >> n;
		string s; cin >> s;
		for (int i = 1; i <= n; ++i)
		{
			string t; cin >> t;
			sam.last = 0;
			for (int j = 0, len = t.length(); j < len; ++j)
				sam.insert(t[j] - 'a');
		}
		long long ans1 = sam.solve();
		sam.last = 0;
		for (int j = 0, len = s.length(); j < len; ++j)
			sam.insert(s[j] - 'a');
		long long ans2 = sam.solve();
		cout << "Case " << cas << ": " << ans2 - ans1 << endl;
		
	}	
}