[BZOJ3626] [LNOI2014]LCA

传送门

http://www.lydsy.com/JudgeOnline/problem.php?id=3626

题目大意

给定一棵树，询问∑bi=adep[lca(i,c)]\sum_{i=a}^bdep[lca(i,c)]

题解

ORZ

我们先考虑可以怎么求两个点的lcalca的深度,我们对其中一个到根上的所有点都+1,然后查询另一个点到根的所有点权和即可

然后对于本题的所有询问[L,R][L,R]都可以拆成[1,L−1]和[1,R][1,L-1]和[1,R]这样我们离线把查询排序,依次把11~nn到根路径都+1然后查询即可

const
maxn=50005;
var
w:array[0..3*maxn,1..2]of longint;
x:array[0..2*maxn,1..4]of longint;
seg:array[0..4*maxn,1..4]of longint;
fa,son,size,dep,top,pos:array[0..maxn]of longint;
ans:array[0..maxn]of int64;
i,j,k:longint;
n,m,len,a,b,c,now:longint;
procedure swap(var a,b:longint);
var c:longint;
begin c:=a; a:=b; b:=c; end;

procedure init(a,b:longint);
begin
w[len,1]:=b;
if w[a,2]=0 then w[a,2]:=len else w[w[a,1],2]:=len;
w[a,1]:=len; inc(len);
end;

procedure sort(l,r:longint);
var i,j,a,b,c,d:longint;
begin
i:=l; j:=r; a:=x[(l+r)div 2,1]; b:=x[(l+r)div 2,2];
repeat
while (x[i,1]<a)or((x[i,1]=a)and(x[i,2]<b)) do inc(i);
while (x[j,1]>a)or((x[j,1]=a)and(x[j,2]>b)) do dec(j);
if not(i>j) then
begin
d:=1; c:=x[i,d]; x[i,d]:=x[j,d]; x[j,d]:=c;
d:=2; c:=x[i,d]; x[i,d]:=x[j,d]; x[j,d]:=c;
d:=3; c:=x[i,d]; x[i,d]:=x[j,d]; x[j,d]:=c;
d:=4; c:=x[i,d]; x[i,d]:=x[j,d]; x[j,d]:=c;
inc(i); dec(j);
end;
until i>j;
if l<j then sort(l,j);
if i<r then sort(i,r);
end;

procedure dfs1(a:longint);
var tt,v:longint;
begin
tt:=w[a,2]; size[a]:=1; v:=0;
while tt<>0 do
begin
if fa[a]<>w[tt,1]
then
begin
fa[w[tt,1]]:=a; dep[w[tt,1]]:=dep[a]+1;
dfs1(w[tt,1]);
inc(size[a],size[w[tt,1]]); if size[w[tt,1]]>size[v] then v:=w[tt,1];
end;
tt:=w[tt,2];
end;
son[a]:=v;
end;

procedure dfs2(a,pre:longint);
var tt:longint;
begin
tt:=w[a,2]; inc(len); pos[a]:=len; top[a]:=pre;
if son[a]<>0 then dfs2(son[a],pre);
while tt<>0 do
begin
if (w[tt,1]<>fa[a])and(w[tt,1]<>son[a]) then dfs2(w[tt,1],w[tt,1]);
tt:=w[tt,2];
end;
end;

procedure build(a,l,r:longint);
var mid:longint;
begin
seg[a,1]:=l; seg[a,2]:=r; seg[a,3]:=0; seg[a,4]:=0;
if l=r then exit;
mid:=(l+r)>>1;
build(a<<1,l,mid); build(a<<1+1,mid+1,r);
end;

procedure pushdown(a:longint);
begin
if seg[a,1]=seg[a,2] then begin seg[a,4]:=0; exit; end;
inc(seg[a<<1,4],seg[a,4]); inc(seg[a<<1,3],seg[a,4]*(seg[a<<1,2]-seg[a<<1,1]+1));
inc(seg[a<<1+1,4],seg[a,4]); inc(seg[a<<1+1,3],seg[a,4]*(seg[a<<1+1,2]-seg[a<<1+1,1]+1));
seg[a,4]:=0;
end;

procedure update(a,l,r:longint);
var mid:longint;
begin
if seg[a,4]<>0 then pushdown(a);
if (l=seg[a,1])and(r=seg[a,2]) then begin inc(seg[a,4]); seg[a,3]:=seg[a,3]+seg[a,2]-seg[a,1]+1; exit; end;
mid:=(seg[a,1]+seg[a,2])>>1;
if r<=mid then update(a<<1,l,r) else
if l>mid then update(a<<1+1,l,r)
else begin update(a<<1,l,mid); update(a<<1+1,mid+1,r); end;
seg[a,3]:=seg[a<<1,3]+seg[a<<1+1,3];
end;

function query(a,l,r:longint):longint;
var mid:longint;
begin
if seg[a,4]<>0 then pushdown(a);
if (l=seg[a,1])and(r=seg[a,2]) then exit(seg[a,3]);
mid:=(seg[a,1]+seg[a,2])>>1;
if r<=mid then exit(query(a<<1,l,r)) else
if l>mid then exit(query(a<<1+1,l,r))
else exit(query(a<<1,l,mid)+query(a<<1+1,mid+1,r));
end;

procedure change(a,b:longint);
begin
while top[a]<>top[b] do
begin
if dep[top[a]]<dep[top[b]] then swap(a,b);
update(1,pos[top[a]],pos[a]);
a:=fa[top[a]];
end;
if dep[a]>dep[b] then swap(a,b);
update(1,pos[a],pos[b]);
end;

function queryans(a,b:longint):int64;
var c:longint; ans:int64;
begin
ans:=0;
while top[a]<>top[b] do
begin
if dep[top[a]]<dep[top[b]] then swap(a,b);
c:=query(1,pos[top[a]],pos[a]); inc(ans,c);
a:=fa[top[a]];
end;
if dep[a]>dep[b] then swap(a,b);
c:=query(1,pos[a],pos[b]); inc(ans,c);
exit(ans);
end;

begin
readln(n,m); len:=n+1;
for i:=2 to n do
begin
readln(a); inc(a);
init(a,i); init(i,a);
end;
len:=0;
for i:=1 to m do
begin
readln(a,b,c); inc(a); inc(b); inc(c);
inc(len); x[len,1]:=a-1; x[len,2]:=c; x[len,3]:=i; x[len,4]:=-1;
inc(len); x[len,1]:=b; x[len,2]:=c; x[len,3]:=i; x[len,4]:=1;
end;
sort(1,len); {x[i,1] x[i,2]} x[0,1]:=len;
build(1,1,n);
dep[1]:=1; fa[1]:=0; size[0]:=0; dfs1(1);
len:=0; dfs2(1,1); now:=0;
for i:=1 to x[0,1] do
begin
for j:=now+1 to x[i,1] do
change(1,j);
now:=x[i,1];
ans[x[i,3]]:=ans[x[i,3]]+queryans(1,x[i,2])*x[i,4];
end;
for i:=1 to m do
writeln(ans[i] mod 201314);
end.