The key of MC in "bzoj3439" KPM

Main topic:

There are n strings, numbered 1~n, that each string in the other string is numbered in a string that is suffixed with the number K small string.

Ideas:

Trie the string upside down, recording the number of each end node (there may be duplicates, so open a vector record). The trie tree is then Dfs ordered, and the sub-tree interval of the end node is recorded. The interval k is small, naturally with the persistence of the segment tree (since the weight is the ID, so it does not need to be discretized). Note: The DFS sequence at the end of the node is encountered only when the serial number of the variable cnt, and the same string of DFS serial number is the same (that is, a tail node may contain many different numbered strings), the left interval takes the first one (such as DFS to x node, numbered 2,x node has two strings, Then their left interval is 2, and the sequence number will be added two times.

Reference from: http://blog.csdn.net/xym_CSDN/article/details/51340321

Code:

1#include <vector>2#include <cstdio>3#include <cstring>4#include <iostream>5 #defineM 5000096 using namespacestd;7Vector <int>Q[m];8 intcnt,trie[m][ -],num[m],com[m], out[m],dat[m],sum[m<<2],lc[m<<2],rc[m<<2];9 CharS[m];Ten  One voidInsintIdChar*str) A { -     intI,l=strlen (str), j=0, K; -      for(i=l-1; i>=0; I--, j=Trie[j][k]) the         if(!trie[j][k=str[i]-'a']) trie[j][k]=++CNT; -num[j]++; q[j].push_back (ID); - } -  + voidDfsintx) - { +     inti; A      for(i=0; i<num[x];i++) com[q[x][i]]=cnt+1; at      for(i=0; i<num[x];i++) dat[++cnt]=Q[x][i]; -      for(i=0;i< -; i++)if(Trie[x][i]) DFS (trie[x][i]); -      for(i=0; i<num[x];i++) out[q[x][i]]=CNT; - } -  - voidBuildintLintRintCurint_cur,intx) in { -sum[cur]=sum[_cur]+1; to     if(L==R)return;intMid=l+r>>1; +     if(X<=mid) lc[cur]=++cnt,rc[cur]=Rc[_cur],build (l,mid,lc[cur],lc[_cur],x); -     ElseLc[cur]=lc[_cur],rc[cur]=++cnt,build (mid+1, r,rc[cur],rc[_cur],x); the } *  $ intAskintLintRintLintRintk)Panax Notoginseng { -     if(L==R)returnL; the     intMid=l+r>>1, t=sum[lc[r]]-Sum[lc[l]]; +     if(t>=k)returnAsk (l,mid,lc[l],lc[r],k); A     Else returnAsk (mid+1, r,rc[l],rc[r],k-t); the } +  - intMain () $ { $     intN,i,x; scanf"%d",&n); -      for(i=1; i<=n;i++) scanf ("%s", s), INS (i,s); -      for(cnt=0, DFS (0), cnt=n+1, i=1; i<=n;i++) Build (1, n,i+1, I,dat[i]); the      for(i=1; i<=n;i++) -     {Wuyiscanf"%d",&x); the         if(sum[ out[i]+1]-SUM[COM[I]]&LT;X) printf ("-1\n"); -         Elseprintf"%d\n", Ask (1, N,com[i], out[i]+1, x)); Wu     } -     return 0; About}

"bzoj3439" KPM the MC cipher key