"BZOJ1717" mode of milk production (suffix array)

Model of "BZOJ1717" milk (suffix array) surface

Permission questions
Hihocoder
Rokua

Exercises

\ (hihocoder\) It's very good.

This question requires the longest overlapping K second son string

The so-called same sub-string
We can understand that if there are two suffixes with the same prefix
Then there is the same substring

If the prefix of the two suffix is the same
So their rankings in \ (sa\) are close

Say it clearly.
If the prefix of the two suffix is the same
Must be the same prefix for a successive suffix in the suffix sort

Therefore, after finding the \ (height\) array
Consider how to calculate the answer:
It's obvious that you can't do it directly.
So two points to the answer

How \ (check\) is there a \ ( k\) repeating substring of length \ (mid\) ?
Since it's a continuous interval,
Therefore, it is necessary to check if there are more than \ ( k\) consecutive \ (height\) \ ( >=mid\)

This is easy to prove:
If you have \ (L.. r\) \ (height\) has exceeded \ (mid\)
So, the proof of any two suffixes in this interval
\ (lcp\) length is at least \ (mid\)
Therefore, this time \ (mid\) must appear more than \ (k\) times

So, vigorously two points can be

#include <iostream>#include <cstdio>#include <cstdlib>#include <cstring>#include <cmath>#include <algorithm>#include <set>#include <map>#include <vector>#include <queue>using namespaceStd#define MAX 1200000inline intRead () {intx=0, t=1;CharCh=getchar (); while((ch<' 0 '|| Ch>' 9 ') &&ch!='-') Ch=getchar ();if(ch=='-') t=-1, Ch=getchar (); while(ch<=' 9 '&&ch>=' 0 ') x=x*Ten+ch-48, Ch=getchar ();returnX*t;}intSa[max],rank[max],x[max],y[max],t[max];intHeight[max],a[max];intN,k;BOOLcmpintIintJintK) {returnY[I]==Y[J]&AMP;&AMP;Y[I+K]==Y[J+K];}voidGetsa () {intm=1000010; for(intI=1; i<=n;++i) t[x[i]=a[i]]++; for(intI=1; i<=m;++i) T[i]+=t[i-1]; for(inti=n;i>=1; i.) sa[t[x[i]]--]=i; for(intk=1; k<=n;k<<=1)    {intp=0; for(intI=1; i<=n;++i) y[i]=0; for(intI=n-k+1; i<=n;++i) y[++p]=i; for(intI=1; i<=n;++i)if(sa[i]>k) Y[++p]=sa[i]-k; for(intI=0; i<=m;++i) t[i]=0; for(intI=1; i<=n;++i) t[x[y[i]]]++; for(intI=1; i<=m;++i) T[i]+=t[i-1]; for(inti=n;i>=1;----) sa[t[x[y[i]]]--]=y[i];        Swap (x, y); x[sa[1]]=p=1; for(intI=2; i<=n;++i) x[sa[i]]=cmp (sa[i],sa[i-1],k)? p:++p;if(p>=n) Break;    M=p; } for(intI=1; i<=n;++i) rank[sa[i]]=i; for(intI=1, j=0; i<=n;++i) {if(j) j--; while(A[i+j]==a[sa[rank[i]-1]+J]) J + +;    Height[rank[i]]=j; }}BOOLCheckinth) {intCnt=0; for(intI=2; i<=n;++i) {if(height[i]0;Elsecnt++;if(cnt==k-1)return true; }return false;}intMain () {n=read (); K=read (); for(intI=1; i<=n;++i) A[i]=read (); Getsa ();intL=1, r=n,ans=0; while(L&LT;=R) {intMid= (l+r) >>1;if(Check (mid)) Ans=mid,l=mid+1;ElseR=mid-1; } printf ("%d\n", ans);return 0;}

"BZOJ1717" mode of milk production (suffix array)