[NOI Online 2021 Improved Group] Building Block Competition-Suffix Automata + Sub-sequence Automata

Synchronized at: https://www.luogu.com.cn/blog/OUYE2020/solution-p7469

answer

Consider the essentially different substrings in the string T at most $n^{2.}$ Violent enumeration judgment is at least$O\left(n^3\right)$, use$hash$ can be optimized to$O\left(n^2\right)$. However, in order to ensure correctness, we do not use$hash$ (Actually, I only used this thing once when doing string questions.

So in order to remove the duplication, the best way is to $T$ string creates a suffix automaton, and traverses the entire suffix automaton by side to be like a$TR\; \&\; ltIE$ tree obtained as information of each sub-string. Due to the suffix automata up to$2n$ points, each point will traverse no more than$n$ times (in fact, more than$n\; is$ much smaller), so the amortized complexity is$O\left(n^2\right)$, where$n>26$ is strictly less than$O(n^2)$ 。

How to take TTSubstring andSS in$T$What about subsequence matching in$S$ ? We greedily record$f_{i,c}$No. iiThe first occurrence of the charactercc after$i$ characters$c$ position, put$i$ 和 $f_{i,c}$Connect the edges between them, and
finally get one at most $n\ast 2$ Trie trie with$6$ sides$trie$ tree. According to greedy thinking, this tree must recordSSInformation about all subsequences of$S$ , this$trie$ tree is called$S$ isa subsequence automaton.

Then we only need to traverse the two automata synchronously, and find the number of all coincident nodes is the answer. This complexity is the smaller value of the traversal times of the two automata, so the total complexity is less than $O(n^2)$ 。

Code

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<cmath>
#include<vector>
#include<queue>
#include<stack>
#define ll long long
#define MAXN 3005
#define uns unsigned
#define INF 0x3f3f3f3f
using namespace std;
inline ll read(){
    
    
	ll x=0;bool f=1;char s=getchar();
	while((s<'0'||s>'9')&&s>0){
    
    if(s=='-')f^=1;s=getchar();}
	while(s>='0'&&s<='9')x=(x<<1)+(x<<3)+s-'0',s=getchar();
	return f?x:-x;
}
struct SAM{
    
    
	int ch[26],len,fa;
	SAM(){
    
    memset(ch,0,sizeof(ch)),len=fa=0;}
}sam[MAXN<<1];
int las=1,tot=1;
inline void samadd(int c){
    
    
	int p=las,np=las=++tot;sam[np].len=sam[p].len+1;
	for(;p&&sam[p].ch[c]==0;p=sam[p].fa)sam[p].ch[c]=np;
	if(!p)sam[np].fa=1;
	else{
    
    int q=sam[p].ch[c];
		if(sam[q].len==sam[p].len+1)sam[np].fa=q;
		else{
    
    
			int nq=++tot;sam[nq]=sam[q],sam[nq].len=sam[p].len+1,sam[q].fa=sam[np].fa=nq;
			for(;p&&sam[p].ch[c]==q;p=sam[p].fa)sam[p].ch[c]=nq;
		}
	}
}
int n,ans;
char a[MAXN],b[MAXN];
int tr[MAXN][26];
inline void dfs(int x,int y){
    
    
	if(!x||!y)return;
	if(x>1)ans++;
	for(int i=0;i<26;i++)
		dfs(sam[x].ch[i],tr[y][i]);
}
signed main()
{
    
    
//	freopen("block.in","r",stdin);
//	freopen("block.out","w",stdout);
	n=read();
	scanf("%s\n%s",a+2,b+1);
	for(int i=n;i>0;i--){
    
    
		for(int j=0;j<26;j++)tr[i][j]=tr[i+1][j];
		tr[i][a[i+1]-'a']=i+1;
	}
	for(int i=1;i<=n;i++)samadd(b[i]-'a');
	dfs(1,1);
	printf("%d\n",ans);
	return 0;
}