Double Stack sort 2008 Noip National League Improvement Group (two-figure dyeing)

Double Stack sorting 

2008 Noip National League Improvement Group

time limit: 1 sspace limit: 128000 KBTopic rank: Master Master
  Title Description Description

Tom has been studying an interesting sort of problem recently. , with 2 stacks of S1 and s2,tom, I want to sort the input sequence in ascending order with the following 4 operation implementations.

Action A

If the input sequence is not empty, the first element is pressed into the stack S1

Action B

If the stack S1 is not empty, eject the top element of the S1 stack to the output sequence

Operation C

If the input sequence is not empty, the first element is pressed into the stack S2

Operation D

If the stack S2 is not empty, eject the top element of the S2 stack to the output sequence

If a 1~n arrangement p can be used to make the output sequence a series of ..., (n-1), N,tom is called P is a "double stack sort arrangement". For example (1,3,2,4) is a "double-stack sort sequence", and (2,3,4,1) is not. Describes an action sequence that sorts (1,3,2,4):<a,c,c,b,a,d,d,b>

Of course, this sequence of operations may have several, for the above example (1,3,2,4),<a,c,c,b,a,d,d,b> is another feasible sequence of operations. Tom wants to know what the sequence of operations with the smallest dictionary order is.

Enter a description input Description

The first line of input is an integer n.

The second line has n spaces separated by a positive integer, constituting a 1~n arrangement.

outputs description output Description

Output a total row, if the input arrangement is not "can be double-stack sort arrangement", the output number 0, otherwise the output dictionary order the smallest sequence of operations, each of the two operations separated by a space, there is no space at the end of the line.

sample input to sample

"Sample 1"

4

1 3 2 4

"Sample 2"

4

2 3 4 1

"Sample 3"

3

2 3 1

Sample output Sample outputs

"Sample 1"

A b A a b b a B

"Sample 2"

0

"Sample 3"

A c a b b d

data size & Hint

30% of the data meet: n<=10

50% of the data meet: n<=50

100% of the data meet: n<=1000

/*If you can not double-stack sequencing, there must be a conflict to the conflict with the edge, because there are only two stacks, determine whether the two-dimensional graph staining can consider the conflict. If there is i<j<k a[k]<a[i]<a[j] must not be a single-stack sort if i<j<k<x have a[x]<a[i]<a[j]<a[k] must not be double-stack sort preprocessing suffix minimum value, Connect the conflict with the edge. */#include<iostream>#include<cstdio>#include<cstring>#include<cstdlib>#include<vector>#include<queue>#include<stack>#defineN 1007#defineINF 0x3f3f3f3f#defineB 1using namespacestd;intn,ans,cnt;intA[n],f[n],col[n],head[n];stack<int>s1,s2;structedge{intu,v,net;} E[n<<1];inlinevoidAddintUintv) {e[++cnt].v=v;e[cnt].net=head[u];head[u]=CNT;} InlineintRead () {intx=0, f=1;CharC=GetChar ();  while(c>'9'|| c<'0'){if(c=='-') f=-1; c=GetChar ();}  while(c>='0'&&c<='9') {x=x*Ten+c-'0'; c=GetChar ();} returnx*F;}voidBFsintu) {Queue<int>q;q.push (u); col[u]=B;  while(!Q.empty ()) {        intnow=Q.front (); Q.pop ();  for(intI=head[now];i;i=e[i].net) {            intv=e[i].v; if(col[v]==-1) col[v]=col[now]^1, Q.push (v); Else if(Col[v]!= (col[now]^1) {printf ("0\n"); exit (0);} }    }}intMain () {memset (col,-1,sizeofcol); N=read ();  for(intI=1; i<=n;i++) a[i]=read (); F[n+1]=inf;  for(inti=n;i>=1; i--) F[i]=min (f[i+1],a[i]);  for(intI=1; i<=n;i++)       for(intj=i+1; j<=n;j++)        if(a[i]>f[j+1] && a[i]<A[j]) Add (i,j), add (j,i);  for(intI=1; i<=n;i++)       if(col[i]==-1) BFS (i); CNT=1;  for(intI=1; i<=n;i++)    {        if(col[i]==b) S1.push (A[i]), printf ("a"); ElseS2.push (A[i]), printf ("C");  while((!s1.empty () && s1.top () ==cnt) | | (!s2.empty () && s2.top () = =CNT)) {            if(!s1.empty () && s1.top () ==cnt) S1.pop (), printf ("b"); ElseS2.pop (), printf ("D"); ++CNT; }    }    return 0;}

Double Stack sort 2008 Noip National League Improvement Group (two-figure dyeing)