luogu6139 【模板】广义后缀自动机（广义SAM）

写了一个离线做法的广义\(\mathrm{SAM}\).

#include<iostream>
#include<string.h>
#include<string>
#include<stdio.h>
#include<algorithm>
#include<vector>
#include<bitset>
#include<math.h>
#include<stack>
#include<queue>
#include<set>
#include<map>
using namespace std;
typedef long long ll;
typedef long double db;
typedef pair<int,int> pii;
const int N=1000000+100;
const db pi=acos(-1.0);
#define lowbit(x) (x)&(-x)
#define sqr(x) (x)*(x)
#define rep(i,a,b) for (register int i=a;i<=b;i++)
#define per(i,a,b) for (register int i=a;i>=b;i--)
#define go(u,i) for (register int i=head[u];i;i=sq[i].nxt)
#define fir first
#define sec second
#define mp make_pair
#define pb push_back
#define maxd 998244353
#define eps 1e-8
inline int read()
{
    int x=0,f=1;char ch=getchar();
    while ((ch<'0') || (ch>'9')) {if (ch=='-') f=-1;ch=getchar();}
    while ((ch>='0') && (ch<='9')) {x=x*10+(ch-'0');ch=getchar();}
    return x*f;
}

namespace My_Math{
	#define N 100000

	int fac[N+100],invfac[N+100];

	int add(int x,int y) {return x+y>=maxd?x+y-maxd:x+y;}
	int dec(int x,int y) {return x<y?x-y+maxd:x-y;}
	int mul(int x,int y) {return 1ll*x*y%maxd;}
	ll qpow(ll x,int y)
	{
		ll ans=1;
		while (y)
		{
			if (y&1) ans=mul(ans,x);
			x=mul(x,x);y>>=1;
		}
		return ans;
	}
	int inv(int x) {return qpow(x,maxd-2);}

	int C(int n,int m)
	{
		if ((n<m) || (n<0) || (m<0)) return 0;
		return mul(mul(fac[n],invfac[m]),invfac[n-m]);
	}

	int math_init()
	{
		fac[0]=invfac[0]=1;
		rep(i,1,N) fac[i]=mul(fac[i-1],i);
		invfac[N]=inv(fac[N]);
		per(i,N-1,1) invfac[i]=mul(invfac[i+1],i+1);
	}
	#undef N
}
using namespace My_Math;

int n;
char s[N];

struct Trie{
	int tot,ch[N][26],fa[N],c[N];
	
	Trie() {tot=1;}
	
	void insert(char *s)
	{
		int now=1,n=strlen(s+1);
		rep(i,1,n)
		{
			int x=s[i]-'a';
			if (!ch[now][x]) {ch[now][x]=(++tot);c[tot]=x;fa[tot]=now;}
			now=ch[now][x];
		}
	}
}tr;

struct Suffix_Automaton{
	int tot,fa[N<<1],ch[N<<1][26],len[N<<1],pos[N<<1];
	queue<int> q;
	
	Suffix_Automaton() {tot=1;}
	
	int insert(int x,int lst)
	{
		int np=(++tot),p=lst;len[np]=len[p]+1;
		while ((p) && (!ch[p][x])) {ch[p][x]=np;p=fa[p];}
		if (!p) {fa[np]=1;return np;}
		int q=ch[p][x];
		if (len[q]==len[p]+1) {fa[np]=q;return np;}
		int nq=(++tot);len[nq]=len[p]+1;
		memcpy(ch[nq],ch[q],sizeof(ch[nq]));
		fa[nq]=fa[q];fa[q]=fa[np]=nq;
		while ((p) && (ch[p][x]==q)) {ch[p][x]=nq;p=fa[p];}
		return np;
	}
	
	void build()
	{
		rep(i,0,25) 
			if (tr.ch[1][i]) q.push(tr.ch[1][i]);
		pos[1]=1;
		while (!q.empty())
		{
			int u=q.front();q.pop();
			//cout << "now " << u << endl;
			pos[u]=insert(tr.c[u],pos[tr.fa[u]]);
			rep(i,0,25)
				if (tr.ch[u][i]) q.push(tr.ch[u][i]);
		}
	}
	
	ll work()
	{
		ll ans=0;
		rep(i,1,tot) ans+=len[i]-len[fa[i]];
		return ans;
	}
}sam;

int main()
{
	n=read();
	rep(i,1,n)
	{
		scanf("%s",s+1);
		tr.insert(s);
	}
	sam.build();
	ll ans=sam.work();
	printf("%lld",ans);
	return 0;
}