HDU3518 suffix array to find the number of different substrings that cannot overlap and recur

Enumerates the length of a substring, grouped by height, if the difference between the SA minimum and SA maximum is greater than the length of the enumeration, this group contributes 1 to the answer.

1#include <iostream>2#include <vector>3#include <algorithm>4#include <string>5#include <string.h>6#include <stdio.h>7#include <queue>8#include <stack>9#include <map>Ten#include <Set> One#include <cmath> A#include <ctime> -#include <cassert> -#include <sstream> the using namespacestd; -  - Const intn=223456; - Const intmod=1e9+7; +  -  + CharS[n]; A structSuffixarray {; at     intSa[n]; -     intT1[n],t2[n],c[n]; -     intRk[n],height[n]; -  -Inlineintcmpint*r,intAintBintl) { -         returnr[a]==r[b]&&r[a+l]==r[b+l]; in     } -     voidCALCSA (Char*s,intNintm) { to         inti,j,p,*x=t1,*y=T2; +          for(i=0; i<m;i++) c[i]=0; -          for(i=0; i<n;i++) c[x[i]=s[i]]++; the          for(i=1; i<m;i++) c[i]+=c[i-1]; *          for(i=n-1; i>=0; i--) sa[--c[x[i]]]=i; $          for(j=1; j<=n;j<<=1){Panax Notoginsengp=0; -              for(i=n-j;i<n;i++) y[p++]=i; the              for(i=0; i<n;i++)if(sa[i]>=j) Y[p++]=sa[i]-j;//rank from small to large, if POS is larger than J, then the second keyword of suffix (sa[i]-j) is P +              for(i=0; i<m;i++) c[i]=0; A              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];//depending on the second keyword from large to small, determine the new round of SA - swap (x, y); $p=1; x[sa[0]]=0; $              for(i=1; i<n;i++) -X[SA[I]]=CMP (y,sa[i-1],sa[i],j)? p1:p + +; -             if(p>=n) Break; them=p; -         }Wuyi     } the     voidCalcheight (Char*s,intN) { -         inti,j,k=0; Wu          for(i=0; i<=n;i++) rk[sa[i]]=i; -          for(i=0; i<n;i++){ About             if(k) k--;//h[i]>=h[i-1]-1 $j=sa[rk[i]-1];//suffix (j) was ranked before suffix (i) -              while(S[i+k]==s[j+k]) k++;//Violent calculation LCP -height[rk[i]]=K; -         } A     } +     intLcpintAintBintLen) { the         if(a==b)returnlen-A; -         intra=rk[a],rb=Rk[b]; $         if(ra>RB) Swap (RA,RB); the         returnQueryst (ra+1, RB); the     } the  the     intst[n][ -];//The second dimension is guaranteed to be greater than log (n<<1) -     voidINITST (intN) { in          for(intI=1; i<=n; i++) thest[i][0]=Height[i]; the          for(intj=1; (1&LT;&LT;J) <=n; J + +) { About             intk=1<< (J-1); the              for(intI=1; i+k<=n; i++) theSt[i][j]=min (st[i][j-1],st[i+k][j-1]); the         } +     } -     intQueryst (intAintb) { the         if(a>b) Swap ( A, a);Bayi         intdis=b-a+1; the         intK=log ((Double) dis)/log (2.0); the         returnMin (st[a][k],st[b-(1&LT;&LT;K) +1][k]); -     } - }suf; the  the intMain () { the      while(SCANF ("%s", s)! =EOF) { the         if(s[0]=='#') Break; -         intn=strlen (s); theSUF.CALCSA (s,n+1, -); the suf.calcheight (s,n); the         //for (int i=0;i<=n;i++) cout<<suf.sa[i]<< "";cout<<endl;94         //for (int i=0;i<=n;i++) cout<<suf.height[i]<< "";cout<<endl; the         intret=0; the          for(intI=1; i<=n/2; i++) { the             intmi=suf.sa[1],ma=suf.sa[1];98              for(intj=2; j<=n;j++) { About                 if(suf.height[j]<i) { -                     if(MI+I&LT;=MA) ret++;101Mi=Suf.sa[j];102Ma=Suf.sa[j];103 104                 } theMi=min (mi,suf.sa[j]);106Ma=Max (ma,suf.sa[j]);107             }108             if(MI+I&LT;=MA) ret++;109         } thecout<<ret<<Endl;111     } the     return 0;113}

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HDU3518 suffix array to find the number of different substrings that cannot overlap and recur