我直接来讲在线好了
这是一个很巧妙的方法，把边作为一个点
做一遍最小生成树，当加如一条边时，我们把这条边两点x,y的并查集的根i,j的父亲都设为这条边代表的点k，由k向i，j连边
这样我们就构建出一棵树，这棵树的叶子都是原来节点
且每棵子树都是在子树根所代表的边的限制下的最小连通块
这样我们就可以通过dfs序（只用对叶子标号）+主席树来维护k大了并通过倍增找到限制
这两题都是一副卡pascal过不了的样子……QAQ
另外网上的一些标称（bzoj3551）似乎没有考虑一个点没有边可走，但询问k=1的情况，可以加数据cha
type node=record
po,next:longint;
end;
way=record
x,y,z:longint;
end;
point=record
l,r,s:longint;
end; var e:array[..] of node;
w:array[..] of way;
v:array[..] of boolean;
anc:array[..,..] of longint;
d,l,r,p,fa,c,a,b,h:array[..] of longint;
tree:array[..*] of point;
num,j,s,size,i,len,t,n,m,q,x,y,ans,k:longint; function min(a,b:longint):longint;
begin
if a>b then exit(b) else exit(a);
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 swap(var a,b:longint);
var c:longint;
begin
c:=a;
a:=b;
b:=c;
end; procedure sort(l,r:longint);
var i,j,x:longint;
begin
i:=l;
j:=r;
x:=a[(l+r) shr ];
repeat
while a[i]<x do inc(i);
while x<a[j] do dec(j);
if i<=j then
begin
swap(a[i],a[j]);
swap(b[i],b[j]);
inc(i);
dec(j);
end;
until i>j;
if l<j then sort(l,j);
if i<r then sort(i,r);
end; procedure qsort(l,r:longint);
var i,j,x:longint;
y:way;
begin
i:=l;
j:=r;
x:=w[(l+r) shr ].z;
repeat
while w[i].z<x do inc(i);
while x<w[j].z do dec(j);
if 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 qsort(l,j);
if i<r then qsort(i,r);
end; procedure dfs(x:longint);
var i,y:longint;
begin
v[x]:=true;
if x<=n then
begin
inc(num);
b[num]:=x;
l[x]:=num;
r[x]:=num;
exit;
end;
i:=p[x];
l[x]:=n;
r[x]:=;
while i<> do
begin
y:=e[i].po;
anc[y,]:=x;
dfs(y);
l[x]:=min(l[x],l[y]);
r[x]:=max(r[x],r[y]);
i:=e[i].next;
end;
end; function find(x,y:longint):longint;
var i:longint;
begin
for i:=size downto do
if a[anc[x,i]]<=y then x:=anc[x,i];
exit(x);
end; function build(l,r:longint):longint;
var m,q:longint;
begin
inc(t);
q:=t;
if l<>r then
begin
m:=(l+r) shr ;
tree[q].l:=build(l,m);
tree[q].r:=build(m+,r);
end;
exit(q);
end; function add(l,r,last,x:longint):longint;
var m,q:longint;
begin
inc(t);
q:=t;
if l=r then tree[q].s:=tree[last].s+
else begin
m:=(l+r) shr ;
if x<=m then
begin
tree[q].r:=tree[last].r;
tree[q].l:=add(l,m,tree[last].l,x);
end
else begin
tree[q].l:=tree[last].l;
tree[q].r:=add(m+,r,tree[last].r,x);
end;
tree[q].s:=tree[tree[q].l].s+tree[tree[q].r].s;
end;
exit(q);
end; function ask(l,r,x,y,k:longint):longint;
var m,s:longint;
begin
if l=r then exit(c[l])
else begin
m:=(l+r) shr ;
s:=tree[tree[y].r].s-tree[tree[x].r].s;
if s>=k then exit(ask(m+,r,tree[x].r,tree[y].r,k))
else exit(ask(l,m,tree[x].l,tree[y].l,k-s));
end;
end; procedure ins(x,y:longint);
begin
inc(len);
e[len].po:=y;
e[len].next:=p[x];
p[x]:=len;
end; begin
readln(n,m,q);
for i:= to n do
begin
read(d[i]);
a[i]:=d[i];
b[i]:=i;
end;
sort(,n);
s:=;
d[b[]]:=;
c[]:=a[];
for i:= to n do
begin
if a[i]<>a[i-] then
begin
inc(s);
c[s]:=a[i];
end;
d[b[i]]:=s;
end;
for i:= to *n do
fa[i]:=i;
for i:= to m do
readln(w[i].x,w[i].y,w[i].z); qsort(,m);
t:=n;
fillchar(a,sizeof(a),);
for i:= to m do
begin
x:=getf(w[i].x);
y:=getf(w[i].y);
if x<>y then
begin
inc(t);
fa[x]:=t;
fa[y]:=t;
a[t]:=w[i].z;
ins(t,x);
ins(t,y);
if t=*n- then break;
end;
end; a[]:=;
len:=t;
for i:= to len do
if not v[i] then dfs(getf(i));
size:=trunc(ln(len)/ln()+0.1);
for j:= to size do
for i:= to len do
begin
x:=anc[i,j-];
anc[i,j]:=anc[x,j-];
end;
t:=;
h[]:=build(,s);
for i:= to num do
h[i]:=add(,s,h[i-],d[b[i]]);
ans:=-;
for i:= to q do
begin
readln(x,y,k);
{ if ans<>-1 then
begin
x:=x xor ans;
y:=y xor ans;
k:=k xor ans;
end; }
x:=find(x,y);
if x<=n then
begin
if k= then ans:=c[d[x]]
else ans:=-;
end
else if tree[h[r[x]]].s-tree[h[l[x]-]].s<k then ans:=-
else ans:=ask(,s,h[l[x]-],h[r[x]],k);
writeln(ans);
end;
end.