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[..,'a'..'z'] of longint;
     f,q,v:array[..] of longint;
     ans:array[..] of string;
     dp:array[..,..,..] of int64;
     t,k,i,j,n,m,l:longint;
     c:char;
     s:string;

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

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

     while h<=r do
     begin
       x:=q[h];
       for c:='a' to 'z' do
         if trie[x,c]> then
         begin
           y:=trie[x,c];
           inc(r);
           q[r]:=y;
           j:=f[x];
           while (j>) and (trie[j,c]=) 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= shl n- then dp[i,p,q]:=
       else dp[i,p,q]:=;
     end;
     if dp[i,p,q]<>- then exit(dp[i,p,q]);
     if (q= shl n-) and (m-i<) then exit(calc(m-i));
     dp[i,p,q]:=;
     for c:='a' to 'z' do
       dp[i,p,q]:=dp[i,p,q]+dfs(i+,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+,go[p,c],q or v[go[p,c]]]> then
         begin
           s[i+]:=c;
           get(i+,go[p,c],q or v[go[p,c]]);
         end;
     end;
   end;

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

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