bzoj1875

我们知道，邻接矩阵自乘t次就是经过t条边的路径数
但这道题显然不能这样，因为不能走回头路
于是我们可以构造边的邻接矩阵矩乘即可

 const mo=;
 type way=record
        po,fr:longint;
      end;
      node=array[..,..] of longint;

 var c,a,w:node;
     ed:array[..] of way;
     x,y,n,m,s,t,e,i,j,ans,len:longint;

 procedure add(x,y:longint);
   begin
     inc(len);
     ed[len].po:=y;
     ed[len].fr:=x;
   end;

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

 procedure quick(n:longint);
   begin
     while n> do
     begin
       if n mod = then mul(a,a,w);
       mul(w,w,w);
       n:=n shr ;
     end;
   end;

 begin
   len:=;
   readln(n,m,t,s,e);
   inc(s);
   inc(e);
   for i:= to m do
   begin
     readln(x,y);
     inc(x); inc(y);
     add(x,y);
     add(y,x);
   end;
   for i:= to len do
   begin
     for j:= to len do
       if (i xor <>j) then
         if ed[i].po=ed[j].fr then w[i,j]:=;
     a[i,i]:=;
     if ed[i].fr=s then c[,i]:=;
   end;
   quick(t-);
   mul(c,c,a);
   for i:= to len do
     if ed[i].po=e then ans:=(ans+c[,i]) mod mo;
   writeln(ans mod mo);
 end.