bzoj1898

这是yhc大牛矩乘论文上的题目，那里面分析得很清楚了
这里就说说我一开始错的地方，我一开始处理每个时刻能通过的邻接矩阵的时候
把以不能访问的点i为起点和终点的都标记为0了，实际上只能标记以i为终点的边即可
很好理解，是自己太脑残了

 const mo=;
 var b,c,g,f:array[..,..] of longint;
     v:array[..,..] of boolean;
     w,d:array[..] of longint;
     a:array[..,..,..] of longint;
     fish,i,j,k,x,y,n,m,s,e,t:longint;

 procedure work(i:longint);
   var j,k:longint;
   begin
     a[i]:=f;
     for j:= to n do
       if v[j,i] then
       begin
         for k:= to n do
           a[i,k,j]:=;
       end;
   end;

 procedure mul(p:longint);
   var i,j,k:longint;
   begin
     b:=c;
     for i:= to n do
       for j:= to n do
       begin
         c[i,j]:=;
         for k:= to n do
           c[i,j]:=(c[i,j]+b[i,k]*a[p,k,j] mod mo) mod mo;
       end;
   end;

 procedure time;
   var i,j,k:longint;
   begin
     for i:= to n do
       for j:= to n do
       begin
         c[i,j]:=;
         for k:= to n do
           c[i,j]:=(c[i,j]+g[i,k]*f[k,j] mod mo) mod mo;
       end;
   end;

 procedure quick(x:longint);
   var i,j:longint;
   begin
     j:=;
     while x> do
     begin
       inc(j);
       d[j]:=x mod ;
       x:=x shr ;
     end;
     fillchar(c,sizeof(c),);
     for i:= to n do
       c[i,i]:=;
     for i:=j downto  do
     begin
       g:=c;
       f:=c;
       time;
       if d[i]= then
       begin
         g:=c;
         f:=b;
         time;
       end;
     end;
   end;

 begin
   readln(n,m,s,e,t);
   inc(s);
   inc(e);
   for i:= to m do
   begin
     readln(x,y);
     inc(x);
     inc(y);
     f[x,y]:=;
     f[y,x]:=;
   end;
   readln(fish);
   for i:= to fish do
   begin
     read(y);
     for j:= to y- do
     begin
       read(x);
       inc(x);
       k:=j;
       while k<= do
       begin
         v[x,k]:=true;
         k:=k+y;
       end;
     end;
   end;
   for i:= to  do
     work(i);
   if t< then
   begin
     c:=a[];
     for i:= to t do
       mul(i);
   end
   else begin
     c:=a[];
     for i:= to  do
       mul(i);
     b:=c;
     quick(t div );
     for i:= to t mod  do
       mul(i);
   end;
   writeln(c[s,e]);
 end.