[2017 山东一轮集训 Day5]LOJ 6071 字符串 - SAM - 拓扑排序

题目大意：给你n个小写字符串$s_1\dots s_n$，问有多少字符串$t$是可接受的。一个串t是可接受的当且仅当存在$t=p_1+\cdots +p_n$，满足$p_i$是$s_i$的子串。$n,\sum |s|\le 10^6$
题解：显然对每个串构造SAM，每个节点若不存在字符c的出边就向后面第一个初始节点有c出边的SAM的的初始节点的c出边连边然后跑一个拓扑排序即可。

#include
#define rep(i,a,b) for(int i=a;i<=b;i++)
#define Rep(i,v) rep(i,0,(int)v.size()-1)
#define lint long long
#define mod 1000000007
#define ull unsigned lint
#define db long double
#define pb push_back
#define mp make_pair
#define fir first
#define sec second
#define gc getchar()
#define debug(x) cerr<<#x<<"="<
#define sp <<" "
#define ln <
using namespace std;
typedef pair<int,int> pii;
typedef set<int>::iterator sit;
inline int inn()
{
	int x,ch;while((ch=gc)<'0'||ch>'9');
	x=ch^'0';while((ch=gc)>='0'&&ch<='9')
		x=(x<<1)+(x<<3)+(ch^'0');return x;
}
const int N=1e6+2e5,T=N*3;
namespace SAM_space{
	int ch[T][26],fa[T],val[T],node_cnt;
	struct SAM{
		int rt,las;
		inline int init() { return las=rt=new_node(0); }
		inline int new_node(int v) { int x=++node_cnt;val[x]=v;return x; }
		inline int extend(int w)
		{
			int p=las,np=new_node(val[p]+1);
			while(p&&!ch[p][w]) ch[p][w]=np,p=fa[p];
			if(!p) fa[np]=rt;
			else{
				int q=ch[p][w],v=val[p]+1;
				if(val[q]==v) fa[np]=q;
				else{
					int nq=new_node(v);
					fa[nq]=fa[q],fa[q]=fa[np]=nq;
					rep(i,0,25) ch[nq][i]=ch[q][i];
					while(p&&ch[p][w]==q) ch[p][w]=nq,p=fa[p];
				}
			}
			return las=np;
		}
	};
}using SAM_space::SAM;SAM s[N];int go[T][26],in[T],vis[T];
inline int add_edge(int x,int y,int i) { return go[x][i]=y,in[y]++; }
inline int build(int x,int n,int *nt)
{
	using SAM_space::ch;if(vis[x]) return 0;vis[x]=1;
	rep(i,0,25)
		if(ch[x][i]) add_edge(x,ch[x][i],i),build(ch[x][i],n,nt);
		else if(nt[i]<=n) add_edge(x,ch[s[nt[i]].rt][i],i);
	return 0;
}
queue<int> q;int dp[T];
inline int solve()
{
	while(!q.empty()) q.pop();dp[1]=1;int ans=0;
	rep(i,1,SAM_space::node_cnt) if(!in[i]) q.push(i);
	while(!q.empty())
	{
		int x=q.front();q.pop();
		ans+=dp[x],(ans>=mod?ans-=mod:0);
		rep(i,0,25)
		{
			int y=go[x][i];dp[y]+=dp[x],
			(dp[y]>=mod?dp[y]-=mod:0);
			if(!(--in[y])) q.push(y);
		}
	}
	return ans;
}
char toc[30];int toi[300];
char str[N];int nxt[N][26],nt[30];
int main()
{
//	freopen("data.in","r",stdin);
	int n=inn();s[0].init();
	rep(i,'a','z') toi[i]=i-'a',toc[i-'a']=char(i);
	rep(i,1,n)
	{
		scanf("%s",str+1);int len=(int)strlen(str+1);s[i].init();
		rep(j,1,len) nxt[i][toi[(int)str[j]]]=i,s[i].extend(toi[(int)str[j]]);
	}
	rep(i,0,25) nxt[n+1][i]=n+1;
	for(int i=n;i>=0;i--) rep(j,0,25) if(!nxt[i][j]) nxt[i][j]=nxt[i+1][j];
	rep(i,0,n) { rep(j,0,25) nt[j]=nxt[i+1][j];build(s[i].rt,n,nt); }
	return !printf("%d\n",solve());
}