bzoj3083 3306

又见bzoj的语言歧视，囧……
bzoj3083过了本地的数据在上面出现各种奇葩的TLE
835083 phile 3083 Time_Limit_Exceed 17092 kb 4872 ms Pascal/Edit 4931B 2015-01-11 19:53:32
10s的时限在逗我？
UPD：现在已经没有问题了……
这两道题目还是不错的
首先这道题是很像动态树的题目，我们发现树的形态不发生变化，而只是根在变化
考虑树链剖分，修改显然跟根的变化毫无关系
主要是查询，首先对于子树的查询肯定是dfs序(而轻重链剖分正是dfs序的一种特殊形式）
下面我们想，一次换跟操作对哪些点的查询会产生影响
画图可知，换根只会影响新根的祖先们的查询，如果不是祖先，那直接查询即可
而是祖先的话，设查询点x和新根路径上离x最近的那个点为y
显然，除去在原图上以y为根的子树，其余部分都应该是换根后的x的子树
由此题目得解，这里给出bzoj3083的代码

 const inf=;
 type node=record
        po,next:longint;
      end;

 var c,e,b,a,d,p,size,top,fa:array[..] of longint;
     w:array[..] of node;
     tree,lazy:array[..*] of longint;
     anc:array[..,..] of longint;
     root,ch,i,s,len,n,m,x,y,z:longint;

 function min(a,b:longint):longint;
   begin
     if a>b then exit(b) else exit(a);
   end;

 procedure push(i:longint);
   begin
     lazy[i*]:=lazy[i];
     lazy[i*+]:=lazy[i];
     tree[i*]:=lazy[i];
     tree[i*+]:=lazy[i];
     lazy[i]:=;
   end;

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

 procedure add(x,y:longint);
   begin
     inc(len);
     w[len].po:=y;
     w[len].next:=p[x];
     p[x]:=len;
   end;

 procedure dfs1(x:longint);
   var i,y:longint;
   begin
     for i:= to s do
     begin
       y:=anc[x,i-];
       if anc[y,i-]= then break;
       anc[x,i]:=anc[y,i-];
     end;
     size[x]:=;
     i:=p[x];
     while i<> do
     begin
       y:=w[i].po;
       if fa[x]<>y then
       begin
         d[y]:=d[x]+;
         fa[y]:=x;
         anc[y,]:=x;
         dfs1(y);
         size[x]:=size[x]+size[y];
       end;
       i:=w[i].next;
     end;
   end;

 procedure dfs2(x:longint);
   var q,i,y:longint;
   begin
     inc(len);
     b[len]:=x;
     c[x]:=len;
     q:=;
     i:=p[x];
     while i<> do
     begin
       y:=w[i].po;
       if fa[y]=x then
         if size[y]>size[q] then q:=y;
       i:=w[i].next;
     end;
     if q<> then
     begin
       top[q]:=top[x];
       dfs2(q);
     end;
     i:=p[x];
     while i<> do
     begin
       y:=w[i].po;
       if (fa[y]=x) and (y<>q) then
       begin
         top[y]:=y;
         dfs2(y);
       end;
       i:=w[i].next;
     end;
     e[x]:=len;   //【c[x],e[x]】表示了以x根的子树所代表的区间
   end;

 procedure build(i,l,r:longint);
   var m:longint;
   begin
     if l=r then tree[i]:=a[b[l]]
     else begin
       m:=(l+r) shr ;
       build(i*,l,m);
       build(i*+,m+,r);
       tree[i]:=min(tree[i*],tree[i*+]);
     end;
   end;

 procedure change(i,l,r,x,y:longint);
   var m:longint;
   begin
     if (x<=l) and (y>=r) then
     begin
       lazy[i]:=z;
       tree[i]:=z;
     end
     else begin
       if lazy[i]<> then push(i);
       m:=(l+r) shr ;
       if x<=m then change(i*,l,m,x,y);
       if y>m then change(i*+,m+,r,x,y);
       tree[i]:=min(tree[i*],tree[i*+]);
     end;
   end;

 procedure work(x,y:longint);
   var f1,f2:longint;
   begin
     f1:=top[x];
     f2:=top[y];
     while f1<>f2 do
     begin
       if d[f1]>=d[f2] then
       begin
         change(,,n,c[f1],c[x]);
         x:=fa[f1];
       end
       else begin
         change(,,n,c[f2],c[y]);
         y:=fa[f2];
       end;
       f1:=top[x];
       f2:=top[y];
     end;
     if c[x]>c[y] then swap(x,y);
     change(,,n,c[x],c[y]);
   end;

 function getans(i,l,r,x,y:longint):longint;
   var m,s:longint;
   begin
     if x>y then exit(inf);
     if (x<=l) and (y>=r) then exit(tree[i])
     else begin
       if lazy[i]<> then push(i);
       m:=(l+r) shr ;
       s:=inf;
       if x<=m then s:=getans(i*,l,m,x,y);
       if y>m then s:=min(s,getans(i*+,m+,r,x,y));
       exit(s);
     end;
   end;

 function check(x,y:longint):boolean;
   var i,k:longint;
   begin
     if d[y]>=d[x] then exit(false);
     for i:=s downto  do
       if d[x]- shl i>=d[y] then x:=anc[x,i];
     if x=y then exit(true) else exit(false);
   end;

 function ask(x:longint):longint;
   var a,z,i:longint;
   begin
     if x=root then exit(tree[]);
     if not check(root,x) then exit(getans(,,n,c[x],e[x]))
     else begin
       z:=root;
       for i:=s downto  do
         if d[z]- shl i>d[x] then z:=anc[z,i];  //找最近点
       a:=min(getans(,,n,,c[z]-),getans(,,n,e[z]+,n));
       exit(a);
     end;
   end;

 begin
   readln(n,m);
   for i:= to n- do
   begin
     readln(x,y);
     add(x,y);
     add(y,x);
   end;
   s:=trunc(ln(n)/ln());
   for i:= to n do
     read(a[i]);
   readln(root);
   dfs1(root);
   top[root]:=root;
   len:=;
   dfs2(root);
   build(,,n);
   for i:= to m do
   begin
     read(ch);
     if ch= then
     begin
       readln(x);
       root:=x;
     end
     else if ch= then
     begin
       readln(x,y,z);
       work(x,y);
     end
     else begin
       readln(x);
       writeln(ask(x));
     end;
   end;
 end.