Large number of dp+ on POJ--1625CENSORED!+AC automatic machine

Topic Link: Click to enter
In fact, it seems that the matrix can be completely done, but because of the use of large numbers, causing memory to open, so with DP write. In fact, the process of DP is still in the trie we use to prohibit the construction of words to walk the process of M-step. We define DP[I][J] to indicate the number of scenarios that reach node J after I step, then the state transition should be dp[i][j]=sum (Dp[i-1][k]), where k represents a node that can walk to J and cannot be a virus node. But in fact this code is not so good to write, in fact, we can use the node J Active to update its sub-node K, so the transfer equation becomes dp[i][next[j][k]]+=dp[i-1][j]. Require next[j][k] cannot make a virus node. All in all, the DP on the AC automaton is to use the trie tree we have built and the above transfer mode for state transfer.

The code is as follows:

#include <iostream>#include <cstring>#include <cstdio>#include <queue>#include <algorithm>#include <map>using namespace STD;// large number of templatesstructbiginteger{inta[ -];enum{MOD =10000}; BigInteger () {memsetA0,sizeof(A)); a[0]=1;}void Set(intx) {memsetA0,sizeof(A)); a[0]=1; a[1]=x;}voidPrint () {printf("%d", a[a[0]]); for(inti=a[0]-1; I>0; i--) {if(a[i]==0){printf("0000");Continue;} for(intk=Ten; k*a[i]<mod; k*=Ten)printf("0");printf("%d", A[i]); }printf("\ n"); }int&operator[] (intP) {returnA[P];}Const int&operator[] (intPConst{returnA[P];} BigIntegeroperator+ (Constbiginteger& B) {BigInteger C; c[0]=max (a[0], b[0]); for(intI=1; i<=c[0]; i++) C[i]+=a[i]+b[i], c[i+1]+=c[i]/mod, C[i]%=mod;if(c[c[0]+1] >0) c[0]++;returnC } BigIntegeroperator* (Constbiginteger& B) {BigInteger C; c[0]=a[0]+b[0]; for(intI=1; i<=a[0]; i++) for(intj=1; j<=b[0]; J + +) {c[i+j-1]+=A[I]*B[J], c[i+j]+=c[i+j-1]/mod, c[i+j-1]%=mod; }if(c[c[0]] ==0) c[0]--;returnC }}one;Const intmaxn=Ten*Ten+5;CharALPH[MAXN]; Map<char,int>HintGetIndex (CharCH) {returnH[CH];}structtrie{intnext[maxn][maxn/2],FAIL[MAXN],FLAG[MAXN];intRoot,l,len;intNewNode () {memset(next[l],-1,sizeof(Next[l])); Flag[l++];returnL-1; }voidInit () {l=0;    Root=newnode (); }voidInsertCharBuf[]) {intlen1=strlen(BUF);intNow=root; for(intI=0; i<len1;i++) {intIndex=getindex (Buf[i]);if(next[now][index]==-1) Next[now][index]=newnode ();        Now=next[now][index]; } flag[now]=1; }voidBuild () { Queue<int>Q; Fail[root]=root; for(intI=0; i<len;i++) {if(next[root][i]==-1) Next[root][i]=root;Else{fail[next[root][i]]=root;          Q.push (Next[root][i]); }        } while(! Q.empty ()) {intNow=q.front ();if(Flag[fail[now]]) flag[now]++; Q.pop (); for(intI=0; i<len;i++) {if(next[now][i]==-1) Next[now][i]=next[fail[now]][i];Else{Fail[next[now][i]]=next[fail[now]][i];//flag[next[now][i]]|=flag[next[fail[now]][i]];Q.push (Next[now][i]); }            }        }    }}; Trie AC;CharSTR[MAXN]; BigInteger DP[MAXN][MAXN];intMain () {intN,m,p;///freopen ("In.txt", "R", stdin);One.Set(1); while(scanf("%d%d%d", &n,&m,&p)!=eof) {h.clear ();scanf('%s ', Alph); for(intI=0;i<strlen(Alph); i++) h[alph[i]]=i;        Ac.init (); ac.len=strlen(Alph); for(intI=0; i<p;i++) {scanf('%s ', str);        Ac.insert (str); } ac.build (); for(intI=0; i<=m;i++) for(intj=0; J<ac. l;j++) Dp[i][j].Set(0); dp[0][0].Set(1); for(intI=1; i<=m;i++) for(intj=0; J<ac. l;j++) for(intk=0; k<n;k++) {if(Ac.flag[ac.next[j][k]])Continue; dp[i][ac.next[j][k]]=dp[i][ac.next[j][k]]+dp[i-1][J];        } BigInteger ans; Ans.Set(0); for(intI=0; I<ac.        l;i++) Ans=ans+dp[m][i];    Ans.print (); }return 0;}

Copyright NOTICE: This article for Bo Master original article, without Bo Master permission not reproduced.

Large number of dp+ on POJ--1625CENSORED!+AC automatic machine