bzoj 2435 BFS

我们可以先将无根树变成有根树，随便选一个点当根就行了，我们选1，bfs求出来每个点 

的size值，代表以它为根节点的子树中有多少个节点，（dfs可能会爆栈），然后再对于每一条

边算就好了

我tle了，也不知道咋会事儿，可能是pascal的int64的运算有些耗时。。。。

 

/**************************************************************

    Problem: 2435

    User: BLADEVIL

    Language: Pascal

    Result: Time_Limit_Exceed

****************************************************************/

 

//By BLADEVIL

var

    n                       :longint;

    pre, other, last, len   :array[0..2000010] of int64;

    l                       :longint;

    flag                    :array[0..1000010] of boolean;

    size, que, dep          :array[0..1000010] of int64;

    ans                     :int64;

    fuse                    :array[0..2000010] of boolean;

     

procedure connect(x,y,z:longint);

begin

    inc(l);

    pre[l]:=last[x];

    last[x]:=l;

    other[l]:=y;

    len[l]:=z;

end;

     

procedure init;

var

    i                       :longint;

    x, y, z                 :longint;

     

begin

    read(n);

    for i:=1 to n-1 do

    begin

        read(x,y,z);

        connect(x,y,z);

        connect(y,x,z);

    end;

end;

 

procedure bfs;

var

    h, t, cur               :longint;

    q, p                    :longint;

begin

    h:=0; t:=1;

    que[1]:=1;

    flag[1]:=true;

    while h<>t do

    begin

        inc(h);

        cur:=que[h];

        q:=last[cur];

        while q<>0 do

        begin

            p:=other[q];

            if not flag[p] then

            begin

                fuse[q]:=true;

                inc(t);

                que[t]:=p;

                dep[p]:=dep[cur]+1;

                flag[p]:=true;

            end;

            q:=pre[q];

        end;

    end;

end;

 

procedure main;

var

    i                       :longint;

    q, p                    :longint;

begin

    bfs;

    for i:=n downto 1 do

    begin

        q:=last[que[i]];

        size[que[i]]:=1;

        while q<>0 do

        begin

            p:=other[q];

            if dep[p]>dep[que[i]] then inc(size[que[i]],size[p]);

            q:=pre[q];

        end;

    end;

    for i:=1 to l do if fuse[i] then ans:=ans+abs(n-2*size[other[i]])*len[i];

    writeln(ans);

end;

 

begin

    init;

    main;

end.