bzoj2631 3282

这两题都是link cut tree的裸题
之前看Qtree的论文，只会在确定父子关系的情况下连边和删边
如果在任意两个点连边删边怎么做呢？
这时候我们不能随意的将一个点的父节点设为另一个点，因为其中某个点的父节点可能已经被设为另外某点了
其实很简单，连边的时候，我们只要把x变成其所在原树的根，这样x是没有父节点了
然后把x的父节点设为y即可，删边、询问路径的道理类似，具体见程序
给出的是bzoj2631的程序

 const mo=;

 var fa,q,a,mul,add,size,sum:array[..] of longint;
     rev:array[..] of boolean;
     son:array[..,..] of longint;
     i,x0,y0,x,y,z,n,m,t:longint;
     ch:char;

 function root(x:longint):boolean;  //判断是Auxiliary tree（splay）上的父节点还是path parent
   begin
     exit((son[fa[x],]<>x) and (son[fa[x],]<>x));
   end;

 procedure swap(var a,b:longint);
   var c:longint;
   begin
     c:=a;
     a:=b;
     b:=c;
   end;

 procedure update(x:longint);
   var l,r:longint;
   begin
     l:=son[x,];
     r:=son[x,];
     size[x]:=(size[l]+size[r]+) mod mo;
     sum[x]:=(a[x]+sum[l]+sum[r]) mod mo;
   end;

 procedure calc(x,c,j:longint);
   begin
     if (c<=) then  //常数优化，少用int64
     begin
       mul[x]:=mul[x]*c mod mo;
       add[x]:=(add[x]*c+j) mod mo;
       sum[x]:=(sum[x]*c+int64(j)*int64(size[x])) mod mo;
       a[x]:=(a[x]*c+j) mod mo;
     end
     else begin
       mul[x]:=int64(mul[x])*int64(c) mod mo;
       add[x]:=(int64(add[x])*int64(c)+j) mod mo;
       sum[x]:=(int64(sum[x])*int64(c)+int64(j)*int64(size[x])) mod mo;
       a[x]:=(int64(a[x])*int64(c)+j) mod mo;
     end;
   end;

 procedure push(x:longint);
   begin
     if rev[x] then
     begin
       rev[son[x,]]:=not rev[son[x,]];
       rev[son[x,]]:=not rev[son[x,]];
       swap(son[x,],son[x,]);
       rev[x]:=false;
     end;
     if (mul[x]<>) or (add[x]<>) then
     begin
       if son[x,]<> then calc(son[x,],mul[x],add[x]);
       if son[x,]<> then calc(son[x,],mul[x],add[x]);
       mul[x]:=;
       add[x]:=;
     end;
   end;

 procedure rotate(x,w:longint);
   var y:longint;
   begin
     y:=fa[x];
     if not root(y) then
     begin
       if son[fa[y],]=y then son[fa[y],]:=x
       else son[fa[y],]:=x;
     end;
     fa[x]:=fa[y];
     son[y,-w]:=son[x,w];
     if son[x,w]<> then fa[son[x,w]]:=y;
     son[x,w]:=y;
     fa[y]:=x;
     update(y);
   end;

 procedure splay(x:longint);
   var y,t,i:longint;
       ff:boolean;
   begin
     t:=;
     i:=x;
     while not root(i) do
     begin
       inc(t);
       q[t]:=i;
       i:=fa[i];
     end;
     inc(t);
     q[t]:=i;
     for i:=t downto  do
       push(q[i]);
     ff:=true;
     if t= then exit;
     while ff do
     begin
       y:=fa[x];
       if y=q[t] then
       begin
         if son[y,]=x then rotate(x,)
         else rotate(x,);
         ff:=false;
       end
       else begin
         if fa[y]=q[t] then ff:=false;
         if son[fa[y],]=y then
         begin
           if son[y,]=x then rotate(y,)
           else rotate(x,);
           rotate(x,);
         end
         else begin
           if son[y,]=x then rotate(x,)
           else rotate(y,);
           rotate(x,);
         end;
       end;
     end;
     update(x);
   end;

 procedure access(x:longint);
   var y:longint;
   begin
     y:=;
     repeat
       splay(x);
       son[x,]:=y;
       update(x);  //这里要记得维护
       y:=x;
       x:=fa[x];
     until x=;
   end;

 procedure makeroot(x:longint);
   begin
     access(x);  //执行access之后，x和当前树的根之间的路径构成Auxiliary tree
     splay(x);   //将x旋到Auxiliary tree的根，这时候x没有右孩子，左子树的点都是x的祖辈
     rev[x]:=not rev[x];  //执行翻转操作，这时候x是Auxiliary tree上最小的点，也就变成了原树的根
   end;

 procedure link(x,y:longint);
   begin
     makeroot(x);
     fa[x]:=y;
   end;

 procedure cut(x,y:longint);
   begin
     makeroot(x);
     access(y);
     splay(y);
     son[y,]:=;
     fa[x]:=;
   end;

 procedure path(x,y:longint); //构成x到y的路径
   begin
     makeroot(x);
     access(y); //x到y路径上的点就在一棵Auxiliary tree中
     splay(y);
   end;

 begin
   readln(n,m);
   for i:= to n do
   begin
     a[i]:=;
     size[i]:=;
     sum[i]:=;
     mul[i]:=;
   end;
   for i:= to n- do
   begin
     readln(x,y);
     link(x,y);
   end;
   for i:= to m do
   begin
     read(ch);
     if ch='+' then
     begin
       readln(x,y,z);
       z:=z mod mo;
       path(x,y);
       calc(y,,z);
     end
     else if ch='-' then
     begin
       readln(x0,y0,x,y);
       cut(x0,y0);
       link(x,y);
     end
     else if ch='*' then
     begin
       readln(x,y,z);
       z:=z mod mo;
       path(x,y);
       calc(y,z,);
     end
     else begin
       readln(x,y);
       path(x,y);
       writeln(sum[y]);
     end;
   end;
 end.