【CF711D】Directed Roads（环，强连通分量）

题意：

　　给一张N个点N条有向边的图，边可以逆向。问任意逆向若干条边使得这张图无环的方案数(mod 1e9+7)。

n<=200000

思路：三个样例给的好 找规律方便很多

易得有N点的环有(2^n)-2中改法，除了不改和都改

剩下的都是链，设除环外还有K个点，他们的总贡献就是2^k，因为都是一条边相连接怎么改也没有环

CF上快速幂要写在外面不然会出现奇奇怪怪的CE

 const mo=;
 var head,vet,next,stack,low,dfn,b,s,flag:array[..]of longint;
     n,tot,i,id,time,top,x,m:longint;
     ans,f,tmp:int64;

 procedure add(a,b:longint);
 begin
  inc(tot);
  next[tot]:=head[a];
  vet[tot]:=b;
  head[a]:=tot;
 end;

 function min(x,y:longint):longint;
 begin
  if x<y then exit(x);
  exit(y);
 end;

 procedure dfs(u:longint);
 var e,v:longint;
 begin
  flag[u]:=;
  inc(top); stack[top]:=u;
  inc(time); dfn[u]:=time; low[u]:=time;
  e:=head[u];
  while e<> do
  begin
   v:=vet[e];
   if flag[v]= then
   begin
    dfs(v);
    low[u]:=min(low[u],low[v]);
   end
    else if s[v]= then low[u]:=min(low[u],low[v]);
   e:=next[e];
  end;
  if dfn[u]=low[u] then
  begin
   inc(id); s[u]:=id; inc(b[id]);
   while (top>)and(stack[top]<>u) do
   begin
    s[stack[top]]:=id;
    inc(b[id]);
    stack[top]:=;
    dec(top);
   end;
   stack[top]:=; dec(top);
  end;
 end;

 begin

  readln(n);
  for i:= to n do
  begin
   read(x);
   add(i,x);
  end;
  for i:= to n do
   if flag[i]= then dfs(i);
  m:=;
  for i:= to n do
   if b[s[i]]= then m:=m+;
  ans:=;
  for i:= to id do
   if b[i]> then
   begin
    f:=; tmp:=;
    while b[i]> do
    begin
     if b[i] mod = then f:=f*tmp mod mo;
     tmp:=tmp*tmp mod mo;
     b[i]:=b[i] div ;
    end;
    ans:=(ans*(f-)) mod mo;
   end;
   ans:=(ans+mo) mod mo;
   f:=; tmp:=;
   while m> do
   begin
    if m mod = then f:=f*tmp mod mo;
    tmp:=tmp*tmp mod mo;
    m:=m div ;
   end;
   ans:=ans*f mod mo;
  writeln(ans);

 end.