bzoj2535 2109

做过4010这题其实就水了

把图反向之后直接拓扑排序做即可，我们可以用链表来优化

每个航班的最小起飞序号就相当于在反向图中不用这个点最迟到哪

 type node=record
        po,next:longint;
      end;

 var e:array[..] of node;
     ans,p,h,suc,du,d,a:array[..] of longint;
     i,x,y,n,m:longint;

 procedure add(x,y:longint);
   begin
     e[i].po:=y;
     e[i].next:=p[x];
     p[x]:=i;
   end;

 procedure merge(x,y:longint);
   begin
     if h[x]= then h[x]:=h[y]
     else begin
       x:=h[x];
       while suc[x]<> do x:=suc[x];
       suc[x]:=h[y];
     end;
   end;

 function min(wh:longint):longint;
   var i,x,y,t,b:longint;
   begin
     for i:= to n do
     begin
       du[i]:=d[i];
       h[i]:=;
     end;
     for i:= to n do
       if (i<>wh) and (d[i]=) then
       begin
         suc[i]:=h[a[i]];
         h[a[i]]:=i;
       end;
     t:=n;
     while t> do
     begin
       x:=h[t];
       if x= then exit(t);
       h[t]:=suc[h[t]];
       i:=p[x];
       while i<> do
       begin
         y:=e[i].po;
         dec(du[y]);
         if (y<>wh) and (du[y]=) then
         begin
           if a[y]<t then b:=a[y]
           else b:=t;
           suc[y]:=h[b];
           h[b]:=y;
         end;
         i:=e[i].next;
       end;
       merge(t-,t);
       dec(t);
     end;
     exit();
   end;

 procedure work1;
   var i,x,y,t,b:longint;
   begin
     for i:= to n do
     begin
       du[i]:=d[i];
       h[i]:=;
     end;
     for i:= to n do
       if d[i]= then
       begin
         suc[i]:=h[a[i]];
         h[a[i]]:=i;
       end;
     t:=n;
     while t> do
     begin
       x:=h[t];
       ans[t]:=x;
       h[t]:=suc[h[t]];
       i:=p[x];
       while i<> do
       begin
         y:=e[i].po;
         dec(du[y]);
         if du[y]= then
         begin
           if a[y]<t then b:=a[y]
           else b:=t;
           suc[y]:=h[b];
           h[b]:=y;
         end;
         i:=e[i].next;
       end;
       merge(t-,t);
       dec(t);
     end;
     for i:= to n- do
       write(ans[i],' ');
     writeln(ans[n]);
   end;

 procedure work2;
   var i:longint;
   begin
     for i:= to n- do
       write(min(i),' ');
     writeln(min(n));
   end;

 begin
   readln(n,m);
   for i:= to n do
   begin
     read(a[i]);
     if a[i]>n then a[i]:=n;
   end;
   for i:= to m do
   begin
     readln(x,y);
     add(y,x);
     inc(d[x]);
   end;
   work1;
   work2;
 end.