其实用不上树链剖分也行，就普通的lca题目
但是经过这道题可以看出树链剖分也可以解决LCA的题目，
注意ptn是dfn的反查，这样能够快速得到dfs序中相应的节点
#include
using namespace std;
typedef long long ll;
typedef pair pii;
const int N = 2e5;
int fa[N],lv[N],siz[N],son[N],top[N],dfn[N],cnt=0;
int ptn[N];
vector G[N];
int n,seg[N<<2];
int query(int s,int t,int l,int r,int p) {
if(s<=l&&r<=t) return seg[p];
int m=(l+r)/2;
if(t<=m) return query(s,t,l,m,p<<1);
if(m+1<=s) return query(s,t,m+1,r,p<<1|1);
return query(s,t,l,m,p<<1)+query(s,t,m+1,r,p<<1|1);
}
void update(int pos,int val,int l,int r,int p){
if(l==r) {
seg[p]=val;
return;
}
int m=(l+r)/2;
if(pos<=m) update(pos,val,l,m,p<<1);
else update(pos,val,m+1,r,p<<1|1);
seg[p]=seg[p<<1]+seg[p<<1|1];
}
int dfs1(int x,int fx) {
son[x]=-1;
siz[x]=1;
fa[x]=fx;
lv[x]=lv[fx]+1;
for(auto p:G[x]) {
int to = p.first;
if(to==fx) continue;
siz[x] += dfs1(to,x);
if(son[x] == -1 || siz[to] > siz[son[x]]) son[x] = to;
}
return siz[x];
}
void dfs2(int x,int root) {
top[x]=root;
dfn[x]=++cnt;
ptn[dfn[x]]=x;
if(son[x]==-1) return;
dfs2(son[x],root);
for(auto p:G[x]) {
int to = p.first;
if(to == son[x] || to == fa[x]) continue;
dfs2(to,to);
}
}
int dist(int x,int y) {
int ans = 0;
while(top[x] != top[y]) {
if(lv[top[x]]dfn[y]) swap(x,y);
ans+=query(dfn[son[x]],dfn[y],1,n,1);
}
return ans;
}
int querylca(int x,int y) {
while(top[x] != top[y]) {
if(lv[top[x]]=k ) {
while(lv[top[x]] > lv[lca]) {
if(lv[x] - lv[top[x]] +1 >= k) break;
k-=lv[x] - lv[top[x]] +1;
x=fa[top[x]];
}
return ptn[dfn[x] -k+1 ];
}
else {
k -= lv[x] - lv[lca] + 1;
k = lv[y] - lv[lca] - k + 1;
while(lv[top[y]] > lv[lca]) {
if(lv[y] - lv[top[y]] + 1 >= k) break;
k -= lv[y] - lv[top[y]] + 1;
y = fa[top[y]];
}
return ptn[dfn[y] - k + 1];
}
}
int a,b,c,k;
void solve() {
memset(seg,0,sizeof(seg));
cnt = 0;
// cin>>n;
scanf("%d",&n);
for(int i=1;i<=n;i++) G[i].clear();
for(int i=1;i>a>>b;
scanf("%d %d",&a,&b);
// cout<>a>>b;
scanf("%d %d %d",&a,&b,&k);
printf("%d\n",kth(a,b,k));
} else break;
}
// for(int i=1;i<=n;i++) cout<