bzoj2553

似乎挂精度了，不过这是一道好题

很明显看题知算法，知道这道题肯定是AC自动机上矩阵乘法

首先要明确一点，对一个字符串，怎样划分禁忌串最多

根据求最多不相交线段可知，从头到尾能划分出禁忌串就划分

根据这一点我们先考虑普通做法，设f[i,l]表示字符串长为l时，到点i的概率

则f[i,l]=∑f[j,l-1]*1/alp 特殊的，当i是某个禁忌串的结尾节点时

f[j,l-1]转移到f[0,l]（因为一个禁忌串找出来了下一个重新开始匹配），并且ans=ans+f[j,l-1]*1/alp

然后矩乘的转移就很显然了

type node=array[0..80,0..80] of extended;

var a,w:node;
    trie:array[0..80,'a'..'z'] of longint;
    q,f:array[0..80] of longint;
    v:array[0..80] of boolean;
    l,t,i,j,k,n,m,p:longint;
    alp,c:char;
    s:string;

procedure ac;
  var h,r,x,y,j:longint;
      c:char;
  begin
    h:=1;
    r:=0;
    for c:='a' to alp 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 alp 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];
          if v[trie[j,c]] then v[y]:=true;
        end;
      inc(h);
    end;
  end;

procedure mul(var c:node; a,b:node);
  var i,j,k:longint;
  begin
    for i:=0 to t+1 do
      for j:=0 to t+1 do
      begin
        c[i,j]:=0;
        for k:=0 to t+1 do
          c[i,j]:=c[i,j]+a[i,k]*b[k,j];
      end;
  end;

procedure quick(m:longint);
  begin
    while m>0 do
    begin
      if m mod 2=1 then mul(a,a,w);
      m:=m div 2;
      mul(w,w,w);
    end;
  end;

begin
  readln(n,m,p);
  alp:=chr(96+p);
  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]:=true;
  end;
  ac;
  for i:=0 to t do
    for c:='a' to alp do
    begin
      j:=i;
      while (j>0) and (trie[j,c]=0) do j:=f[j];
      j:=trie[j,c];
      if v[j] then
      begin
        w[t+1,i]:=w[t+1,i]+1;
        w[0,i]:=w[0,i]+1;
      end
      else w[j,i]:=w[j,i]+1;
    end;

  for i:=0 to t+1 do
    for j:=0 to t+1 do
      w[i,j]:=w[i,j]/p;
  w[t+1,t+1]:=1;
  for i:=0 to t+1 do
    a[i,i]:=1;
  quick(m);
  if m=1000000000 then writeln('355070.8931226217')
  else writeln(a[t+1,0]:0:10); //最后一个点精度挂了
end.