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=10000;
var b,c,g,f:array[0..51,0..51] of longint;
    v:array[0..51,0..51] of boolean;
    w,d:array[0..51] of longint;
    a:array[0..15,0..51,0..51] 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:=1 to n do
      if v[j,i] then
      begin
        for k:=1 to n do
          a[i,k,j]:=0;
      end;
  end;

procedure mul(p:longint);
  var i,j,k:longint;
  begin
    b:=c;
    for i:=1 to n do
      for j:=1 to n do
      begin
        c[i,j]:=0;
        for k:=1 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:=1 to n do
      for j:=1 to n do
      begin
        c[i,j]:=0;
        for k:=1 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:=0;
    while x>0 do
    begin
      inc(j);
      d[j]:=x mod 2;
      x:=x shr 1;
    end;
    fillchar(c,sizeof(c),0);
    for i:=1 to n do
      c[i,i]:=1;
    for i:=j downto 1 do
    begin
      g:=c;
      f:=c;
      time;
      if d[i]=1 then
      begin
        g:=c;
        f:=b;
        time;
      end;
    end;
  end;

begin
  readln(n,m,s,e,t);
  inc(s);
  inc(e);
  for i:=1 to m do
  begin
    readln(x,y);
    inc(x);
    inc(y);
    f[x,y]:=1;
    f[y,x]:=1;
  end;
  readln(fish);
  for i:=1 to fish do
  begin
    read(y);
    for j:=0 to y-1 do
    begin
      read(x);
      inc(x);
      k:=j;
      while k<=12 do
      begin
        v[x,k]:=true;
        k:=k+y;
      end;
    end;
  end;
  for i:=0 to 12 do
    work(i);
  if t<12 then
  begin
    c:=a[1];
    for i:=2 to t do
      mul(i);
  end
  else begin
    c:=a[1];
    for i:=2 to 12 do
      mul(i);
    b:=c;
    quick(t div 12);
    for i:=1 to t mod 12 do
      mul(i);
  end;
  writeln(c[s,e]);
end.