bzoj4009

这是一道神题，首先我们不难先到整体二分吧

下面的问题就是，求出对于每个水果，有多少盘子是他的子路径

直接考虑不是很容易，我们换个思路，考虑对于每个盘子，哪些水果能包含它

我们假设盘子a,b，dep[a]<dep[b]

如果b在a的子树内，不难发现，水果路径必须是一个点在b的子树内（包括b），一个在c的子树外（不包括c，c是ab路径上a下一个点）

否则 ，水果路径必须两个点分别在a,b子树内

子树问题不难想到dfs序，设子树x区间为l[x],r[x],盘子路径x,y(l[x]<l[y])满足条件为

则要满足x∈[1,l[c]-1],y∈[l[b],r[b]或x∈[l[b],r[b]], y∈[l[c]+1,n] 当b在a的子树内

否则x∈[l[a],r[a]],y∈[l[b],r[b]，把(x,y)看做一个点

也就是现在询问每个点被多少矩形覆盖（注意第一种情形两个矩形是不相交的，所以覆盖的矩形数目就是盘子数目

矩形覆盖的经典做法是把矩形(x0,y0),(x1,y1)看成4个带权值的点

(x0,y0) 1, (x0,y1+1) -1, (x1+1,y0) -1 (x1+1,y1+1) 1，

这样每个询问点最下方（包括边界）的点权值和就是覆盖询问点的矩形数目（画图可知）

而这个问题我们对一维排序，一维树状数组即可快速解决

 type node=record
        po,next:longint;
      end;
      point=record
        x,y,id,z:longint;
      end;

 var e:array[..] of node;
     dep,ans,st,a,b,c,p,d:array[..] of longint;
     anc:array[..,..] of longint;
     q,w,tw:array[..] of point;
     can,v:array[..] of boolean;
     j,tot,t,x,y,z,i,len,n,m,k,xx:longint;

 function lowbit(x:longint):longint;
   begin
     exit(x and (-x));
   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);
     e[len].po:=y;
     e[len].next:=p[x];
     p[x]:=len;
   end;

 procedure dfs(x:longint);
   var i,y:longint;
   begin
     inc(t);
     a[x]:=t;
     i:=p[x];
     while i<> do
     begin
       y:=e[i].po;
       if a[y]= then
       begin
         anc[y,]:=x;
         dep[y]:=dep[x]+;
         dfs(y);
       end;
       i:=e[i].next;
     end;
     b[x]:=t;
   end;

 function jump(x,d:longint):longint;
   var i:longint;
   begin
     for i:= downto  do
       if dep[x]- shl i>=d then x:=anc[x,i];
     exit(x);
   end;

 function cmp(a,b:point):boolean;
   begin
     exit((a.x<b.x) or (a.x=b.x) and (a.y<b.y));
   end;

 procedure sortc(l,r:longint);
   var i,j,x:longint;
   begin
     i:=l;
     j:=r;
     x:=c[(l+r) shr ];
     repeat
       while c[i]<x do inc(i);
       while x<c[j] do dec(j);
       if not(i>j) then
       begin
         swap(c[i],c[j]);
         inc(i);
         dec(j);
       end;
     until i>j;
     if l<j then sortc(l,j);
     if i<r then sortc(i,r);
   end;

 procedure sortw(l,r:longint);
   var i,j:longint;
       x,y:point;
   begin
     i:=l;
     j:=r;
     x:=w[(l+r) shr ];
     repeat
       while cmp(w[i],x) do inc(i);
       while cmp(x,w[j]) do dec(j);
       if not(i>j) then
       begin
         y:=w[i]; w[i]:=w[j]; w[j]:=y;
         inc(i);
         dec(j);
       end;
     until i>j;
     if l<j then sortw(l,j);
     if i<r then sortw(i,r);
   end;

 procedure sortq(l,r:longint);
   var i,j:longint;
       x,y:point;
   begin
     i:=l;
     j:=r;
     x:=q[(l+r) shr ];
     repeat
       while cmp(q[i],x) do inc(i);
       while cmp(x,q[j]) do dec(j);
       if not(i>j) then
       begin
         y:=q[i]; q[i]:=q[j]; q[j]:=y;
         inc(i);
         dec(j);
       end;
     until i>j;
     if l<j then sortq(l,j);
     if i<r then sortq(i,r);
   end;

 procedure open(x0,x1,y0,y1,z:longint);
   begin
     inc(x1); inc(y1);
     inc(t); w[t].x:=x0; w[t].y:=y0; w[t].id:=; w[t].z:=z;
     inc(t); w[t].x:=x0; w[t].y:=y1; w[t].id:=-; w[t].z:=z;
     inc(t); w[t].x:=x1; w[t].y:=y0; w[t].id:=-; w[t].z:=z;
     inc(t); w[t].x:=x1; w[t].y:=y1; w[t].id:=; w[t].z:=z;
   end;

 procedure work(x,w:longint);
   begin
     while x<=n+ do
     begin
       if not v[x] then
       begin
         v[x]:=true;
         inc(tot);
         st[tot]:=x;
       end;
       d[x]:=d[x]+w;
       x:=x+lowbit(x);
     end;
   end;

 function ask(x:longint):longint;
   begin
     ask:=;
     while x> do
     begin
       ask:=ask+d[x];
       x:=x-lowbit(x);
     end;
   end;

 procedure cdq(opx,opy,qx,qy,l,r:longint);
   var m,t,i,j,s,f1,f2,z,q1,q2:longint;
   begin
     if (l>r) or (qx>qy) then exit;
     if l=r then
     begin
       for i:=qx to qy do
         ans[q[i].id]:=c[l];
       exit;
     end;
     m:=(l+r) shr ;  //二分答案
     t:=;
     for i:=opx to opy do
       if w[i].z<=c[m] then
       begin
         inc(t);
         tw[t]:=w[i];
       end;

     z:=;
     j:=;
     for i:=qx to qy do
     begin
       can[i]:=false;
       while (j<=t) and (tw[j].x<=q[i].x) do  //扫描询问
       begin
         work(tw[j].y,tw[j].id);
         inc(j);
       end;
       s:=ask(q[i].y);
       if s>=q[i].z then
       begin
         ans[q[i].id]:=c[m];
         inc(z);
         can[i]:=true;
       end
       else q[i].z:=q[i].z-s;
     end;
     f1:=opx-; f2:=opx+t-;  //划分操作区间
     for i:=opx to opy do
       if w[i].z<=c[m] then
       begin
         inc(f1);
         tw[f1]:=w[i];
       end
       else begin
         inc(f2);
         tw[f2]:=w[i];
       end;
     for i:=opx to opy do
       w[i]:=tw[i];
     q1:=qx-; q2:=qx+z-;
     for i:=qx to qy do
       if can[i] then
       begin
         inc(q1);
         tw[q1]:=q[i];
       end
       else begin
         inc(q2);
         tw[q2]:=q[i];
       end;
     for i:=qx to qy do
       q[i]:=tw[i];
     for i:= to tot do
     begin
       s:=st[i];
       v[s]:=false;
       d[s]:=;
     end;
     tot:=;
     cdq(opx,f1,qx,q1,l,m);
     cdq(f1+,opy,q1+,qy,m+,r);
   end;

 begin
   readln(n,m,k);
   for i:= to n- do
   begin
     readln(x,y);
     add(x,y);
     add(y,x);
   end;
   dfs();
   for j:= to trunc(ln(n)/ln())+ do
     for i:= to n do
     begin
       x:=anc[i,j-];
       anc[i,j]:=anc[x,j-];
     end;

   t:=;
   for i:= to m do
   begin
     readln(x,y,z);
     if a[x]>a[y] then swap(x,y);
     if a[y]<=b[x] then
     begin
       xx:=jump(y,dep[x]+);
       if a[xx]<> then open(,a[xx]-,a[y],b[y],z);
       if b[xx]<n then open(a[y],b[y],b[xx]+,n,z);
     end
     else open(a[x],b[x],a[y],b[y],z);
     c[i]:=z;
   end;
   sortc(,m);
   sortw(,t);
   for i:= to k do
   begin
     readln(x,y,z);
     if a[x]>a[y] then swap(x,y);
     q[i].x:=a[x]; q[i].y:=a[y]; q[i].id:=i; q[i].z:=z;
   end;
   sortq(,k);
   cdq(,t,,k,,m);
   for i:= to k do
     writeln(ans[i]);
 end.