bzoj1559

自动机上状压dp，把单词是否存在压成二进制位
注意这里面某些单词会包含其他单词，所以某些自动机上有些状态点对应多个二进制位
方案只要再顺着有方案的状态搜一遍即可

var trie,go:array[0..110,'a'..'z'] of longint;
    f,q,v:array[0..110] of longint;
    ans:array[0..50] of string;
    dp:array[0..25,0..100,0..1023] of int64;
    t,k,i,j,n,m,l:longint;
    c:char;
    s:string;

function calc(n:longint):int64;
  var i:longint;
  begin
    calc:=1;
    for i:=1 to n do
      calc:=calc*26;
  end;

procedure ac;
  var h,r,x,y,j,i:longint;
      c:char;
  begin
    h:=1;
    r:=0;
    for c:='a' to 'z' do
      if trie[0,c]>0 then
      begin
        inc(r);
        q[r]:=trie[0,c];
      end;

    while h<=r do
    begin
      x:=q[h];
      for c:='a' to 'z' do
        if trie[x,c]>0 then
        begin
          y:=trie[x,c];
          inc(r);
          q[r]:=y;
          j:=f[x];
          while (j>0) and (trie[j,c]=0) do j:=f[j];
          f[y]:=trie[j,c];
          v[y]:=v[y] or v[trie[j,c]];
        end;
      inc(h);
    end;
  end;

function dfs(i,p,q:longint):int64;
  var c:char;
  begin
    if i=m then
    begin
      if q=1 shl n-1 then dp[i,p,q]:=1
      else dp[i,p,q]:=0;
    end;
    if dp[i,p,q]<>-1 then exit(dp[i,p,q]);
    if (q=1 shl n-1) and (m-i<2) then exit(calc(m-i));
    dp[i,p,q]:=0;
    for c:='a' to 'z' do
      dp[i,p,q]:=dp[i,p,q]+dfs(i+1,go[p,c],q or v[go[p,c]]);
    exit(dp[i,p,q]);
  end;

procedure get(i,p,q:longint);
  var c:char;
      j:longint;
  begin
    if i=m then
    begin
      inc(t);
      ans[t]:=s;
    end
    else begin
      for c:='a' to 'z' do
        if dp[i+1,go[p,c],q or v[go[p,c]]]>0 then
        begin
          s[i+1]:=c;
          get(i+1,go[p,c],q or v[go[p,c]]);
        end;
    end;
  end;

begin
  readln(m,n);
  for i:=1 to n do
  begin
    j:=0;
    readln(s);
    l:=length(s);
    for k:=1 to l do
    begin
      if trie[j,s[k]]=0 then
      begin
        inc(t);
        trie[j,s[k]]:=t;
      end;
      j:=trie[j,s[k]];
    end;
    v[j]:=1 shl (i-1);
  end;
  ac;
  for i:=0 to t do
    for c:='a' to 'z' do
    begin
      j:=i;
      while (j>0) and (trie[j,c]=0) do j:=f[j];
      go[i,c]:=trie[j,c];
    end;

  fillchar(dp,sizeof(dp),255);
  writeln(dfs(0,0,0));
  if dp[0,0,0]<=42 then
  begin
    t:=0;
    s:='';
    for i:=1 to m do
      s:=s+' ';
    get(0,0,0);
    for i:=1 to t do
      writeln(ans[i]);
  end;
end.