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[0..10010] of node;
    ans,p,h,suc,du,d,a:array[0..2010] 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]=0 then h[x]:=h[y]
    else begin
      x:=h[x];
      while suc[x]<>0 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:=1 to n do
    begin
      du[i]:=d[i];
      h[i]:=0;
    end;
    for i:=1 to n do
      if (i<>wh) and (d[i]=0) then
      begin
        suc[i]:=h[a[i]];
        h[a[i]]:=i;
      end;
    t:=n;
    while t>0 do
    begin
      x:=h[t];
      if x=0 then exit(t);
      h[t]:=suc[h[t]];
      i:=p[x];
      while i<>0 do
      begin
        y:=e[i].po;
        dec(du[y]);
        if (y<>wh) and (du[y]=0) 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-1,t);
      dec(t);
    end;
    exit(0);
  end;

procedure work1;
  var i,x,y,t,b:longint;
  begin
    for i:=1 to n do
    begin
      du[i]:=d[i];
      h[i]:=0;
    end;
    for i:=1 to n do
      if d[i]=0 then
      begin
        suc[i]:=h[a[i]];
        h[a[i]]:=i;
      end;
    t:=n;
    while t>0 do
    begin
      x:=h[t];
      ans[t]:=x;
      h[t]:=suc[h[t]];
      i:=p[x];
      while i<>0 do
      begin
        y:=e[i].po;
        dec(du[y]);
        if du[y]=0 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-1,t);
      dec(t);
    end;
    for i:=1 to n-1 do
      write(ans[i],' ');
    writeln(ans[n]);
  end;

procedure work2;
  var i:longint;
  begin
    for i:=1 to n-1 do
      write(min(i),' ');
    writeln(min(n));
  end;

begin
  readln(n,m);
  for i:=1 to n do
  begin
    read(a[i]);
    if a[i]>n then a[i]:=n;
  end;
  for i:=1 to m do
  begin
    readln(x,y);
    add(y,x);
    inc(d[x]);
  end;
  work1;
  work2;
end.