Poj 2774 Long long Message (suffix array)

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Poj 2774 Long Long Message

Title Description:

Give a A, a, two string to find the longest common substring?

Problem Solving Ideas:

The longest common substring of a A, B string can be converted to the longest common prefix of an array of string suffixes. Concatenate b strings behind a string, separating the R strings with ' $ '. Find the height of the R string array, the largest height[i] (satisfies sa[i] and sa[i-1] is not in the same text string is the answer. The time complexity is probably O (| a|+| b|) *log (| a|+| b|)). (Parade holiday used to engage in this, feeling still a little strange feeling, but adhering to the previous learning style, first write down the template, later in the water problem slowly realized good)

1#include <cstdio>2#include <cstring>3#include <iostream>4#include <algorithm>5 using namespacestd;6typedefLong LongLL;7 Const intMAXN =200010;8 9 intSA[MAXN], RANK[MAXN], HEIGHT[MAXN];Ten intT1[MAXN], T2[MAXN], R[MAXN], C[MAXN]; One BOOLCMP (int*STR,intAintBintk) A{//rank adjacent to the two strings, the first second key words are the same, rank -     returnSTR[A]==STR[B] && str[a+k]==str[b+K]; - } the intDa (int*STR,intNintm) - { -     int*x = T1, *y =T2, I, J; -n + +; +     //Cardinal Sort, -      for(i=0; i<m; i++) C[i] =0; +      for(i=0; i<n; i++) C[x[i] = Str[i]] + +; A      for(i=1; i<m; i++) C[i] + = c[i-1]; at      for(i=n-1; i>=0; i--) sa[--C[str[i]] =i; -      for(j=1; j<=n; j*=2) -     { -         //sort the second keyword with an SA array -         intp =0; -          for(i=n-j; i<n; i++) y[p++] = i;//You cannot add a string of length J, and the second keyword is the smallest in          for(i=0; i<n; i++)if(Sa[i] >= j) y[p++] = Sa[i]-J; -         //enumeration rank, a string with a starting point greater than J, to multiply to         //Update SA array +          for(i=0; i<m; i++) C[i] =0; -          for(i=0; i<n; i++) C[x[y[i]] + +; the          for(i=1; i<m; i++) C[i] + = c[i-1]; *          for(i=n-1; i>=0; i--) sa[--C[x[y[i]] =Y[i]; $         //Update x ArrayPanax Notoginseng swap (x, y); -p =1, x[sa[0]] =0; the          for(i=1; i<n; i++) +X[sa[i]] = cmp (y, sa[i-1], Sa[i], j)? p1:p + +; A         if(P >=N) the              Break; +m =p; -     } $     //Calculate rank Array $      for(i=0; i<n; i++) -Rank[sa[i]] =i; -     //Calculating the height array theN--; -     intK =0;Wuyi      for(i=0; i<n; i++) the{//a point of lifting, height[rank] -         if(k) K--; Wuj = sa[rank[i]-1]; -          while(Str[i+k]==str[j+k]) k++; AboutHeight[rank[i]] =K; $     } - } - intMain () - { A     Charstr1[maxn/2], str2[maxn/2]; +      while(SCANF ("%s%s", STR1, str2)! =EOF) the     { -         intN1 =strlen (str1); $         intN2 =strlen (str2); the         intn = n1 + N2 +1; the          for(intI=0; i<n; i++) the         { the             if(I <N1) -R[i] =Str1[i]; in             Else if(i = =N1) theR[i] ='$'; the             Else AboutR[i] = str2[i-n1-1]; the         } theR[n] =0; theDa (r, N, -); +         intAns =0; -          for(intI=1; i<n; i++) the{//i is rankBayi             if(sa[i]<n1&&sa[i-1]>n1 | | sa[i]>n1&&sa[i-1]<N1) the                 //rank adjacent suffix string not in the same strings theAns =Max (ans, height[i]); -         } -printf ("%d\n", ans); the     } the     return 0; the}

Poj 2774 Long long Message (suffix array)