[BZOJ2946] [Poi2000] Common string problem solving report | suffix automata

Because Sam wants to be concise ... So I wrote the question again ...

But it's good to learn something new ...

　　here 's how to use SA to do this problem.

First, as with two strings, build a Sam for the first string

And then each string runs on this Sam and matches it.

But what is the answer to our final question?

is a state substring that has the largest minimum value of the matching length in all strings

And then for each string

We can record its maximum matching length on each state substring.

Finally, a very critical shift is needed.

is to update the node pointed to by the fail pointer with the value of the current node

For example, this situation

If one matches to the left three nodes at a time, match to the two nodes on the right (two matches in different strings)

So obviously, the common substring length of these two strings is 2 is present

But since we haven't moved, the point pointed to by the fail pointer does not store the previous information and then it goes wrong.

　　

And then as for the order of transfer, we can follow the depth sequence

This can be achieved with a bucket row.

1 Program bzoj2946;2 ConstMAXN =100010;3 varN,i,j,now,maxl,root,c,tot,cnt,t:longint;4s:array[1..5]of ansistring;5mx,fail,q,b,ans,tem:array[-1.. Maxn]of Longint;6a:array[-1.. maxn,-1.. -]of Longint;7 8 function Max (a,b:longint): Longint;9 beginTen     ifA>b then exit (a)Elseexit (b); One end; A  - function min (a,b:longint): Longint; - begin the     ifA<b then exit (a)Elseexit (b); - end; -  - function Insert (p,c:longint): Longint; + varNp,q,nq:longint; - begin +Inc (CNT); np:=cnt;mx[np]:=mx[p]+1; A      while(p<>0) and (a[p,c]=0) Do at begin -a[p,c]:=np;p:=Fail[p]; - end; -     ifp=0Then Fail[np]:=rootElse - begin -q:=A[p,c]; in         ifmx[q]=mx[p]+1Then fail[np]:=qElse - begin toInc (CNT); nq:=cnt;mx[nq]:=mx[p]+1; +a[nq]:=A[q]; -fail[nq]:=Fail[q]; thefail[np]:=nq;fail[q]:=NQ; *              whileA[p,c]=q Do $ beginPanax Notoginsenga[p,c]:=nq;p:=Fail[p]; - end; the end; + end; A exit (NP); the end; +  - begin $ READLN (n); $      fori:=1to n Doreadln (S[i]); -cnt:=1; root:=1; t:=Root; -      fori:=1To Length (s[1]) DoT:=insert (T,ord (s[1, I])- the); the      fori:=2To CNT Doans[i]:=Mx[i]; -Fillchar (b,sizeof(b),0);Wuyi      fori:=2To CNT DoInc (B[mx[i]); the      fori:=2To CNT DoInc (b[i],b[i-1]); -      fori:=2To CNT Doans[i]:=Mx[i]; Wu      forI:=cnt Downto2  Do - begin About Dec (b[mx[i]]); $q[b[mx[i]]]:=i; - end; -      fori:=1to n Do - begin Anow:=root;maxl:=0; +Fillchar (TEM,sizeof(TEM),0); the          forj:=1To Length (S[i]) Do - begin $C:=ord (S[i][j])- the; the             ifA[now,c]<>0Then begin Now:=a[now,c];inc (MAXL); endElse the begin the                  while(now<>0) and (a[now,c]=0) Donow:=Fail[now]; the                 ifnow=0Then BEGIN now:=root;maxl:=0; endElseBegin maxl:=mx[now]+1; now:=A[now,c];end; - end; intem[now]:=Max (TEM[NOW],MAXL); the end; the          forJ:=cnt Downto0  Dotem[fail[q[j]]]:=Max (Tem[fail[q[j]]],tem[q[j]]); About          forj:=2To CNT Doans[j]:=min (ans[j],tem[j]); the end; thetot:=0; the      fori:=2To CNT Do ifAns[i]>tot then tot:=Ans[i]; + writeln (tot); -End.

Compare a bit. Code reduced by One-third, space reduced by nine-tenths ... It's a lot easier to write.

Sam Dafa Good

　　

05/. May

[bzoj2946][poi2000] Common string problem Solving report | suffix automaton