HDU3341 Lost ' s Revenge (AC automaton +DP)

The title is to rearrange a DNA to contain the most number-theoretic genes.

Considering that memory is probably the only way to represent state:

Dp[i][a][c][g][t], which represents the number of arithmetical genes with the largest number of ACGT and the node with the current suffix status of Automaton I

Where ACGT can be hashed into an integer (a*c*g*t+c*g*t+g*t+t), so that the two-dimensional array on the line, and the second dimension at most 11*11*11*11.

The next shift is still me for everyone, I was thrown into a queue to update the status value with the queue.

This problem is quite card constant, had to hand-written queue, the last 4500msAC, or almost timed out, the code is also very messy.

1#include <cstdio>2#include <cstring>3 using namespacestd;4 inttn,ch[555][4],fail[555],flag[555];5 intidx[ -];6 voidInsertChar*s) {7     intx=0;8      for(intI=0; T[n]; ++i) {9         inty=Idx[s[i]];Ten         if(ch[x][y]==0) ch[x][y]=++tn; Onex=Ch[x][y]; A     } -++Flag[x]; - } the intque[555]; - voidinit () { -memset (fail,0,sizeof(fail)); -     intFront=0, rear=0; +      for(intI=0; i<4; ++i) { -         if(ch[0][i]) que[rear++]=ch[0][i]; +     } A      while(front!=rear) { at         intx=que[front++]; -          for(intI=0; i<4; ++i) { -             if(Ch[x][i]) que[rear++]=ch[x][i],fail[ch[x][i]]=ch[fail[x]][i],flag[ch[x][i]]+=Flag[ch[fail[x]][i]]; -             Elsech[x][i]=Ch[fail[x]][i]; -         } -     } in } - intd[555][14641]; to intquex[555*14641],quey[555*14641]; + intMain () { -idx['A']=0; idx['C']=1; idx['G']=2; idx['T']=3; the     Charstr[ -]; *     intN,cse=0; $      while(~SCANF ("%d", &n) &&N) {Panax Notoginsengtn=0; -memset (CH,0,sizeof(CH)); thememset (Flag,0,sizeof(flag)); +          while(n--){ Ascanf"%s", str); the Insert (str); +         } - init (); $scanf"%s", str); $         inttimes[4]={0}; -          for(intI=0; Str[i]; ++i) { -++Times[idx[str[i]]; the         } -         intTcal0= (times[1]+1) * (times[2]+1) * (times[3]+1);Wuyi         intTcal1= (times[2]+1) * (times[3]+1); the         intTcal2= (times[3]+1); -memset (d,-1,sizeof(d)); Wud[0][0]=0; -         intFront=0, rear=0, x,y,ny,cnt[4]; Aboutquex[rear]=0; quey[rear]=0; ++Rear; $          while(front!=rear) { -X=quex[front]; Y=quey[front]; ++Front; -ny=y; -cnt[0]=ny/tcal0; Any-=cnt[0]*tcal0; +cnt[1]=ny/Tcal1; theny-=cnt[1]*Tcal1; -cnt[2]=ny/Tcal2; $cnt[3]=ny-cnt[2]*Tcal2; the              for(intI=0; i<4; ++i) { the                 if(Cnt[i]>=times[i])Continue; the++Cnt[i]; theny=cnt[0]*tcal0+cnt[1]*tcal1+cnt[2]*tcal2+cnt[3]; -                 if(d[ch[x][i]][ny]==-1){ ind[ch[x][i]][ny]=d[x][y]+Flag[ch[x][i]]; theQuex[rear]=ch[x][i]; Quey[rear]=ny; ++Rear; the}Else if(d[ch[x][i]][ny]<d[x][y]+Flag[ch[x][i]]) { Aboutd[ch[x][i]][ny]=d[x][y]+Flag[ch[x][i]]; the                 } the--Cnt[i]; the             } +         } -y=times[0]*tcal0+times[1]*tcal1+times[2]*tcal2+times[3]; the         intres=0;Bayi          for(intx=0; x<=tn; ++x) { the             if(Res<d[x][y]) res=D[x][y]; the         } -printf"Case %d:%d\n",++cse,res); -     } the     return 0; the}

HDU3341 Lost ' s Revenge (AC automaton +DP)