bzoj2085

首先看到k的范围就该知道这题不是倍增就是矩乘

首先肯定要求出任意一对串(a,b) a的后缀与b的前缀相同的最长长度是多少

考虑到kmp求出的失配指针是一个串最长后缀和前缀相等的长度

这里多个串我们只要用ac自动机即可

具体的，我们只要建立自动机，然后记录每个状态点是哪些串的子串

然后我们只要从每个串的结尾节点出发，顺着失配指针统计即可

然后我们把每个串看作一个点，点之间的边长度就是虽代表串的后缀前缀相同的最长长度

这个问题等价于求经过k条边的最短长度，倍增轻松搞定

 const inf=*;
 type node=record
        po,next:longint;
      end;
      mtx=array[..,..] of int64;

 var lcp:array[..,..] of longint;
     d,p,w,fa,q:array[..] of longint;
     loc:array[..] of longint;
     e:array[..*] of node;
     trie:array[..,'a'..'z'] of longint;
     f,a:mtx;
     ss:ansistring;
     t,len,i,n,m,j,k:longint;
     ans:int64;

 function min(a,b:int64):int64;
   begin
     if a>b then exit(b) else exit(a);
   end;

 function max(a,b:longint):longint;
   begin
     if a>b then exit(a) else exit(b);
   end;

 procedure add(x,y:longint);
   begin
     inc(len);
     e[len].po:=y;
     e[len].next:=p[x];
     p[x]:=len;
   end;

 procedure ac;
   var h,r,x,y,j: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:=fa[x];
           while (j>) and (trie[j,c]=) do j:=fa[j];
           fa[y]:=trie[j,c];
         end;
       inc(h);
     end;
   end;

 procedure get(k,x:longint);
   var i,j:longint;
   begin
     while fa[x]<> do
     begin
       x:=fa[x];
       i:=p[x];
       while i<> do
       begin
         j:=e[i].po;
         lcp[k,j]:=max(lcp[k,j],d[x]);
         i:=e[i].next;
       end;
     end;
   end;

 procedure mul(var c:mtx; a,b:mtx);
   var i,j,k:longint;
   begin
     for i:= to n do
       for j:= to n do
         c[i,j]:=inf;

     for k:= to n do
       for i:= to n do
         for j:= to n do
           c[i,j]:=min(c[i,j],a[i,k]+b[k,j]);
   end;

 begin
   readln(n,m);
   for i:= to n do
   begin
     readln(ss);
     w[i]:=length(ss);
     j:=;
     for k:= to w[i] do
     begin
       if trie[j,ss[k]]= then
       begin
         inc(t);
         trie[j,ss[k]]:=t;
       end;
       j:=trie[j,ss[k]];
       d[j]:=k;
       add(j,i);
     end;
     loc[i]:=j;
   end;
   ac;
   for i:= to n do
     get(i,loc[i]);
   for i:= to n do
     for j:= to n do
       a[i,j]:=w[j]-lcp[i,j];

   for i:= to n do
     for j:= to n do
       if i<>j then f[i,j]:=inf;
   dec(m);
   while m> do
   begin
     if m mod = then mul(f,f,a);
     m:=m shr ;
     mul(a,a,a);
   end;
   ans:=inf;
   for i:= to n do
     for j:= to n do
       ans:=min(ans,f[i,j]+w[i]);
   writeln(ans);
 end.