Bzoj 4566 Jzyzoj 1547 [haoi2016t5] Find the same substring suffix array

http://172.20.6.3/Problem_Show.asp?id=1547

http://www.lydsy.com/JudgeOnline/problem.php?id=4566

It seems that the suffix is a positive solution, but the suffix array + and check set can also mess with a off, define the string size of integer variables when char siz cause re or something; I'm probably a zz.

By the way, save a board, copy Purple Xuan learn elder sister's board.

1#include <iostream>2#include <cstdio>3#include <cstring>4#include <algorithm>5#include <cmath>6 using namespacestd;7 Const intmaxn=200010;8 Const intPl= -;9 intsa[maxn*2+pl]={};//Number I is an array starting from sa[i]Ten intrk[maxn*2+pl]={};//Ranking of I One intheight[maxn*2+pl]={};//The longest same prefix length in rank i-1 as the rank no. i A inttemp[maxn*2+pl]={};//Temporary rankings - intcnt[maxn*2+pl]={};//the number of prefixes (prefixes and) of the Type I (dictionary order) prefix - intp[maxn*2+pl]={};//temporary storage of the last SA, processing the suffix length of different cases.  the Charch[maxn*2+pl]={},ch1[maxn]={},ch2[maxn]={}; - intSiz1,siz2,siz; - intfa[maxn*2+pl]={},a[maxn*2+pl]={},lef[maxn*2+pl]={},rig[maxn*2+pl]={}; -InlineBOOLEquintXintYintL) {returnrk[x]==rk[y]&&rk[x+l]==rk[y+l];} + voidSA () { -      for(intI=1; i<=siz;i++) rk[i]=ch[i],sa[i]=i; +      for(inti,sig=255, l=0, pos=0;p Os<siz;sig=pos) {//L starting from 0 is pretreatment Apos=0; at          for(i=siz-l+1; i<=siz;i++) p[++pos]=i; -          for(i=1; i<=siz;i++)if(sa[i]>l) p[++pos]=sa[i]-l; -          for(i=0; i<=sig;i++) cnt[i]=0; -          for(i=1; i<=siz;i++) cnt[rk[p[i]]]++; -          for(i=1; i<=sig;i++) cnt[i]+=cnt[i-1]; -          for(i=siz;i>0; i--) {sa[cnt[rk[p[i]]]]=p[i];cnt[rk[p[i]]]--;} inpos=0; -          for(i=1; i<=siz;i++){ to             if(Equ (sa[i],sa[i-1],l)) temp[sa[i]]=Pos; +             Elsetemp[sa[i]]=++Pos; -} for(i=1; i<=siz;i++) rk[i]=Temp[i]; the         if(l==0) l=1; *         Elsel<<=1; $     }Panax Notoginseng      for(intI=1, k=0; i<=siz;i++){ -         /*for each suffix of a position, the suffix of the next position can match the shortest length the must be greater than or equal to the length at which the position can be matched-1, obviously. So it's an O (n) algorithm. +         */ A         if(rk[i]==1) {k=0;Continue;} the         if(k>0) k--; +         intj=sa[rk[i]-1]; -          while(Ch[i+k]==ch[j+k]) k++; $height[rk[i]]=K; $     } - } - BOOLMCMP (intXintY) {returnHeight[x]>height[y];} the intGETFA (intx) { -     if(X!=fa[x]) fa[x]=GETFA (fa[x]);Wuyi     returnFa[x]; the } - intMain () { Wu     //freopen ("a.in", "R", stdin); -scanf"%s", &ch1); siz1=strlen (CH1); Aboutscanf"%s", &AMP;CH2); siz2=strlen (CH2); $ch[siz1+1]='Z'+1; siz=siz1+siz2+1; -      for(intI=0; i<siz1;i++) ch[i+1]=Ch1[i]; -      for(intI=0; i<siz2;i++) ch[siz1+2+i]=Ch2[i]; - SA (); A      for(intI=1; i<=siz;i++){ +a[i]=fa[i]=i; theLef[i]= (sa[i]<=siz1); -rig[i]=1^Lef[i]; $}sort (A +1, A +1+siz,mcmp); the     Long Longans=0; the     intx, y; the      for(intI=1; i<=siz;i++){ the         if(a[i]==1)Continue; -X=GETFA (A[i]); Y=GETFA (a[i]-1); inans+= (Long Long) height[a[i]]* (Long Long) (lef[x]*rig[y]+rig[x]*lef[y]); thelef[x]+=lef[y];rig[x]+=rig[y];fa[y]=x; the}printf ("%lld\n", ans); About     return 0; the}

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Bzoj 4566 Jzyzoj 1547 [haoi2016t5] Find the same substring suffix array