bzoj 1180 LCT

给出n个结点以及每个点初始时对应的权值wi。起始时点与点之间没有连边。有3类操作： 1、bridge A B：询问结点A与结点B是否连通。如果是则输出“no”。否则输出“yes”，并且在结点A和结点B之间连一条无向边。 2、penguins A X：将结点A对应的权值wA修改为X。 3、excursion A B：如果结点A和结点B不连通，则输出“impossible”。否则输出结点A到结点B的路径上的点对应的权值的和。给出q个操作，要求在线处理所有操作。数据范围：1<=n<=30000, 1<=q<=300000, 0<=wi<=1000。

LCT模板题...

var        n,m,x,y         :longint;        i               :longint;        ch              :char;        s               :string;        son             :array[-1..30010,0..1] of longint;        father,sum      :array[-1..30010] of longint;        flag            :array[-1..30010] of boolean;        w               :array[-1..30010] of longint;procedure swap(var a,b:longint);var        c:longint;begin   c:=a; a:=b; b:=c;end;procedure update(x:longint);begin   sum[x]:=sum[son[x,0]]+sum[son[x,1]]+w[x];end;function root(x:longint):boolean;begin   if x=son[father[x],0] then exit(false);   if x=son[father[x],1] then exit(false);   exit(true);end;procedure renew(x:longint);begin   flag[x]:=not flag[x];end;procedure push_down(x:longint);begin   if flag[x] then   begin      swap(son[x,0],son[x,1]);      if son[x,0]<>-1 then renew(son[x,0]);      if son[x,1]<>-1 then renew(son[x,1]);   end;   flag[x]:=false;end;procedure push(x:longint);begin   if not root(x) then push(father[x]);   push_down(x);end;procedure ro(x,y:longint);var        f:longint;begin   f:=father[x];   if son[x,y xor 1]<>-1 then father[son[x,y xor 1]]:=f;   son[f,y]:=son[x,y xor 1];   if f=son[father[f],0] then son[father[f],0]:=x else     if f=son[father[f],1] then son[father[f],1]:=x;   father[x]:=father[f];   father[f]:=x;   son[x,y xor 1]:=f;   update(f);   update(x);end;procedure splay(x:longint);var        u,v:longint;begin   while not root(x) do   begin      if root(father[x]) then ro(x,ord(x=son[father[x],1])) else      begin         if x=son[father[x],0] then u:=-1 else u:=1;         if father[x]=son[father[father[x]],0] then v:=-1 else v:=1;         if u*v=1 then         begin            ro(father[x],ord(x=son[father[x],1]));            ro(x,ord(x=son[father[x],1]));         end else         begin            ro(x,ord(x=son[father[x],1]));            ro(x,ord(x=son[father[x],1]));         end;      end;   end;   update(x);end;procedure access(x:longint);var        y:longint;begin   y:=-1;   while x<>0 do   begin      push(x);      splay(x);      son[x,1]:=y;      update(x);      y:=x;      x:=father[x];   end;end;procedure link(x,y:longint);begin   access(x); splay(x); renew(x);   access(y); splay(y);   son[y,1]:=x;   father[x]:=y;   update(y);end;function find_root(x:longint):longint;begin   access(x); splay(x);   while son[x,0]<>-1 do x:=son[x,0];   exit(x);end;begin   read(n);   for i:=1 to n do read(w[i]);   for i:=1 to n do sum[i]:=w[i];   fillchar(son,sizeof(son),255);   readln(m);   for i:=1 to m do   begin      read(ch); s:='';      while(ch<>' ') do      begin         s:=s+ch;         read(ch);      end;      if s='bridge' then      begin         readln(x,y);         if find_root(x)=find_root(y) then writeln('no') else         begin            writeln('yes');            link(x,y);         end;      end else      if s='penguins' then      begin         readln(x,y);         access(x); splay(x); renew(x);         w[x]:=y; update(x);      end else      begin         readln(x,y);         if find_root(x)<>find_root(y) then writeln('impossible') else         begin            access(x); splay(x); renew(x);            access(y); splay(y);            writeln(sum[y]);         end;      end;   end;end.

——by Eirlys