【GDOI2017模拟2.14】A

Description

Input

Sample Input

5 5 4
3 4 2 3 3
2 4
5 1
1 3
2 1
2 5 2
6 7 0
7 7 0
2 4 2

Sample Output

0 0 3
0 3 1
0 0 1
1 2 0

Data Constraint

注意：Q<=300000

Hint

题解

一开始想的是用一些神奇的离线方法做，想了半天之后发现这题是强制在线的QAQ
然后就看了一发题解，结果题解只有：主席树入门题这几个字(。・・)ノ
想了一会儿之后想到可以用权值线段树对每个节点维护它到根节点路径上点权值的分布情况
于是就发现了用主席树的方法

首先我们用主席树预处理每一个节点到根节点路径的点权值的情况，然后每一次询问一组(x,y)，我们就可以随便倍增一下，然后主席树快速算出两点之间的点的分布情况就可以了

贴代码

var
    tree:array[0..6000005,1..3]of longint;
    a,b:array[0..600005,1..2]of longint;
    bz:array[0..300005]of boolean;
    cc,root,fa,deep:array[0..300005]of longint;
    ff:array[0..300005,0..19]of longint;
    i,j,k,l,m,n,x,y,z,p,last,tmp,t,xx,yy,q:longint;
    tt:array[0..3]of longint;
procedure qsort(l,r:longint);
var
    i,j,mid:longint;
begin
    i:=l;
    j:=r;
    mid:=a[(i+j) div 2,1];
    repeat
        while a[i,1]<mid do inc(i);
        while a[j,1]>mid do dec(j);
        if i<=j then
        begin
            a[0]:=a[i];
            a[i]:=a[j];
            a[j]:=a[0];
            inc(i);
            dec(j);
        end;
    until i>j;
    if i<r then qsort(i,r);
    if l<j then qsort(l,j);
end;
procedure star;
begin
    b[a[1,1],1]:=1;
    for i:=2 to n*2-2 do
    if a[i,1]<>a[i-1,1] then
    begin
        b[a[i-1,1],2]:=i-1;
        b[a[i,1],1]:=i;
    end;
    b[a[i,1],2]:=i;
end;
procedure make_ff;
var
    i,j:longint;
begin
    for j:=1 to 18 do
        for i:=1 to n do ff[i,j]:=ff[ff[i,j-1],j-1];
end;
procedure maketree(var num,x:longint;l,r:longint);
var
    mid:longint;
begin
    tree[p]:=tree[x];
    inc(tree[p,3]);
    x:=p;
    inc(p);
    if l=r then exit;
    mid:=(l+r) div 2;
    if num<=mid then maketree(num,tree[x,1],l,mid)
        else maketree(num,tree[x,2],mid+1,r);
end;
procedure find(v1,v2,l,r,x,y,z:longint);
var
    mid:longint;
begin
    if (l=x) and (r=y) then
        tt[z]:=tt[z]+tree[v2,3]-tree[v1,3]
    else
    begin
        mid:=(l+r) div 2;
        if y<=mid then find(tree[v1,1],tree[v2,1],l,mid,x,y,z) else
        if x>mid then find(tree[v1,2],tree[v2,2],mid+1,r,x,y,z) else
        begin
            find(tree[v1,1],tree[v2,1],l,mid,x,mid,z);
            find(tree[v1,2],tree[v2,2],mid+1,r,mid+1,y,z);
        end;
    end;
end;
procedure dfs(x:longint);
var
    i:longint;
begin
    bz[x]:=true;
    for i:=b[x,1] to b[x,2] do
    if (i<>0) and (bz[a[i,2]]=false) then
    begin
        fa[a[i,2]]:=x;
        ff[a[i,2],0]:=x;
        deep[a[i,2]]:=deep[x]+1;
        root[a[i,2]]:=root[x];
        maketree(cc[a[i,2]],root[a[i,2]],1,m);
        dfs(a[i,2]);
    end;
end;
begin
    //assign(input,'jzoj4969.in'); reset(input);
    assign(input,'a.in'); reset(input);
    assign(output,'a.out'); rewrite(output);
    readln(n,m,q);
    for i:=1 to n do read(cc[i]);
    readln;
    for i:=1 to n-1 do
    begin
        readln(a[i,1],a[i,2]);
        a[n+i-1,1]:=a[i,2];
        a[n+i-1,2]:=a[i,1];
    end;
    qsort(1,n*2-2);
    star;
    p:=1;
    maketree(cc[1],root[1],1,m);
    deep[1]:=1;
    dfs(1);
    make_ff;
    for i:=1 to q do
    begin
        readln(x,y,z);
        x:=x xor last;
        y:=y xor last;
        z:=z xor last;
        xx:=x;
        yy:=y;
        if deep[y]>deep[x] then
        begin
            tmp:=x;
            x:=y;
            y:=tmp;
        end;
        k:=deep[x];
        for j:=18 downto 0 do
        if k-1<<j>=deep[y] then
        begin
            k:=k-1<<j;
            x:=ff[x,j];
        end;
        for j:=18 downto 0 do
        if ff[x,j]<>ff[y,j] then
        begin
            x:=ff[x,j];
            y:=ff[y,j];
        end;
        if x<>y then x:=ff[x,0];
        fillchar(tt,sizeof(tt),0);
        if z>1 then
        begin
            find(root[x],root[xx],1,m,1,z-1,1);
            find(root[x],root[yy],1,m,1,z-1,1);
        end;
        find(root[x],root[xx],1,m,z,z,2);
        find(root[x],root[yy],1,m,z,z,2);
        if z<m then
        begin
            find(root[x],root[xx],1,m,z+1,m,3);
            find(root[x],root[yy],1,m,z+1,m,3);
        end;
        if cc[x]<z then inc(tt[1]) else
        if cc[x]=z then inc(tt[2]) else
        inc(tt[3]);
        last:=tt[1] xor tt[2];
        last:=last xor tt[3];
        writeln(tt[1],' ',tt[2],' ',tt[3]);
    end;
    close(input); close(output);
end.