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;}
}