Suffix Array code detailed

That's a long-learned suffix array.

Thought is still easy to understand

The biggest problem is

Code can not understand

And then, in the process of constantly simulating research,

Finally figured out a little

Just decided to write it down.

Or you'll forget qaq.

Here is the code

PS: First of all to understand the Cardinal sort

definition:c[] Array: base-ordered buckets

X[] Array: Similar to rank array, holds the rank of the string that is currently sorted at the beginning of each position

Sa[] Array: sa[i] Save the first position of the string beginning of the rank I

Y[] Array: Second keyword sort, y[i] save the first position of the string beginning of the rank I

1  voidBUILDSA (intm) {2       int*x = t, *y = t2;//This is the sort of single character3        for(inti =0; I < m; i++) C[i] =0;//emptying the C array4        for(inti =0; I < n; i++) C[x[i] = s[i]]++;//initial length is 1 o'clock, x[i] = = S[i]5        for(inti =1; I < m; i++) C[i] + = c[i-1];//Get each location6        for(inti = n-1; I >=0; i--) Sa[--c[x[i]] = i;//ranked C[x[i]], the character is from the I position, each record ranked 1, to the previous position, to ensure that the ranking and the corresponding position is not duplicated7        for(intK =1; K <= N; K <<=1) {8        　　 intp =0;//For counting purposes.9        　　  for(inti = n-k; I < n; i++) y[p++] = i;//0 is the smallest, put the back need to fill 0 in front of the firstTen        　　  for(inti =0; I < n; i++)if(Sa[i] >= k) y[p++] = sa[i]-K;//Sa[i] < K is not the second keyword, sa[i] 's ranking is incremented sequentially, so read directly into the Y array, minus the part that is not the second keyword before One        　　  for(inti =0; I < m; i++) C[i] =0;//emptying the C array A        　　  for(inti =0; I < n; i++) c[x[y[i]]]++;//At this point in Y has been sorted by the second key word, X is the first key word, and then directly the base of the first keyword can be sorted -           for(inti =1; I < m; i++) C[i] + = c[i-1]; -           for(inti = n-1; I >=0; i--) sa[--c[x[y[i] []] =Y[i]; the swap (x, y); -         //discretization Ranking -p =1; x[sa[0]] =0; -            for(inti =1; I < n; i++) +X[sa[i]] = y[sa[i-1] [= Y[sa[i]] && y[sa[i-1] + K] = = Y[sa[i] + K]? P-1: p++;//Two keywords are the same and ranked the same -          if(P >= N) Break;//P represents the number of different rankings, if the number has reached N, then the order is complete, you can directly exit +m =p; A      } at}

This is the practice of getting an SA array

However

An SA array is no use

Here is the algorithm for getting the height array

Specific proof can be referenced

Https://wenku.baidu.com/view/ed1be61e10a6f524ccbf85fd.html

This paper

It's very well written.

Just no comment ...

First, it's the code.

definition:rank[] Array: Opposite to sa[] array, rank[i] = J represents the rank of suffix starting with I

1 voidgetheight () {2     intK =0;3      for(inti =0; I < n; i++) Rank[sa[i] =i;4      for(inti =0; I < n; i++)    {5         if(k) k--;//Because H[i] >= h[i-1]-1 (h[i] read the definition of the paper), so K needs-16         intj = Sa[rank[i]-1]; Get the position where the previous suffix begins7          while(S[i + K] = = S[j + K]) k++;//Compare from the back of K, until it is not the same8Height[rank[i]] =K;9     }Ten}

Suffix Array code detailed