拓扑序列的最小字典序列

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考场上又写挂的一道签到题。。。

我们发现这题要求得到一个最小字典序列

显然找到所有序列然后排序是不可取的，

那么我们不能使用平常的拓扑排序方法，怎么搞使得在每次处理拓扑顺序的时候来维护呢？

用小根堆维护入度为0的点即可，输入的时候统计入度

{$inline on}

const maxn=100010; type node=record pos,next:longint; end; var heap,getin,head,ans:array[0..maxn] of longint; edge:array[0..maxn*2] of node; n,m,cnt,top:longint; check:boolean; procedure add(u,v:longint); inline; begin inc(cnt); edge[cnt].pos:=v; edge[cnt].next:=head[u]; head[u]:=cnt; end; procedure swap(var a,b:longint); inline; begin a:=a xor b; b:=a xor b; a:=a xor b; end; procedure up(i:longint); inline; var j:longint; begin j:=i>>1; while j>0 do

    begin

        if heap[i]<heap[j] then

            begin swap(heap[i],heap[j]); i:=j; j:=j>>1; end

        else break; end; end; procedure down(i:longint); inline; var j:longint; begin j:=i<<1; while j<=top do

        begin

            if (j<top) and (heap[j]>heap[j+1]) then inc(j); if heap[i]>heap[j] then

                begin swap(heap[i],heap[j]); i:=j; j:=j<<1; end

            else break; end; end; procedure init; var i,x,y:longint; begin read(n,m); for i:=1 to m do

        begin read(x,y); add(x,y); inc(getin[y]); end; for i:=1 to n do

        if getin[i]=0 then

            begin inc(top); heap[top]:=i; end; check:=true; end; procedure main; var i,j,x,y:longint; begin

    for j:=1 to n do

        begin

            if (top=0) then

                begin check:=false; break; end; x:=heap[1]; ans[j]:=x; swap(heap[1],heap[top]); dec(top); down(1); i:=head[x]; while i<>0 do

                begin y:=edge[i].pos; dec(getin[y]); if getin[y]=0 then

                        begin inc(top); heap[top]:=y; up(top); end; i:=edge[i].next; end; end; end; procedure print; var i:longint; begin

    if not check then writeln('OMG.') else 

        begin

            for i:=1 to n-1 do write(ans[i],' '); writeln(ans[n]); end; end; begin init; main; print; end.

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