bzoj2333

好题，先离线把连通块变成连续的区间
每次连通块合并就相当于两个区间合并
这样就轻易的用线段树解决了

 type node=record
        wh:string[];
        x,y:longint;
      end;

 var lazy,tree:array[..*] of longint;
     q:array[..] of node;
     a,b,c,last,next,fa:array[..] of longint;
     i,n,m,x,y,t:longint;
     ch:char;

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

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

 function getf(x:longint):longint;
   begin
     if fa[x]<>x then fa[x]:=getf(fa[x]);
     exit(fa[x]);
   end;

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

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

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

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

 begin
   readln(n);
   for i:= to n do
   begin
     read(a[i]);
     fa[i]:=i;
     last[i]:=i;
   end;
   readln(m);
   for i:= to m do
   begin
     read(ch);
     q[i].wh:=ch;
     read(ch);
     q[i].wh:=q[i].wh+ch;
     if q[i].wh='U ' then
     begin
       readln(q[i].x,q[i].y);
       x:=getf(q[i].x);
       y:=getf(q[i].y);
       if x=y then continue;
       fa[y]:=x;
       next[last[x]]:=y;
       last[x]:=last[y];
     end
     else if (q[i].wh='A1') or (q[i].wh='A2') then
       readln(q[i].x,q[i].y)
     else if (q[i].wh='F3') then readln
     else readln(q[i].x);
   end;

   for i:= to n do
     if fa[i]=i then
     begin
       x:=i;
       while x<> do
       begin
         inc(t);
         b[x]:=t;
         c[t]:=x;
         x:=next[x];
       end;
     end;

   build(,,n);
   for i:= to n do
   begin
     fa[i]:=i;
     last[i]:=i;
   end;

   for i:= to m do
     if q[i].wh='U ' then
     begin
       x:=getf(q[i].x);
       y:=getf(q[i].y);
       if x=y then continue;
       fa[y]:=x;
       last[x]:=last[y];
     end
     else if q[i].wh='A1' then
       add(,,n,b[q[i].x],b[q[i].x],q[i].y)
     else if q[i].wh='A2' then
     begin
       x:=getf(q[i].x);
       y:=last[x];
       add(,,n,b[x],b[y],q[i].y);
     end
     else if q[i].wh='A3' then
     begin
       inc(tree[],q[i].x);
       inc(lazy[],q[i].x);
     end
     else if q[i].wh='F1' then
       writeln(ask(,,n,b[q[i].x],b[q[i].x]))
     else if q[i].wh='F2' then
     begin
       x:=getf(q[i].x);
       y:=last[x];
       writeln(ask(,,n,b[x],b[y]));
     end
     else writeln(tree[]);
 end.