LCA,即在图论和计算机科学中，最近公共祖先（英語：lowest common ancestor）是指在一个树或者有向无环图中同时拥有v和w作为后代的最深的节点。在这里，我们定义一个节点也是其自己的后代，因此如果v是w的后代，那么w就是v和w的最近公共祖先。　　——wikipedia

# E. 1-Trees and Queries

## Description

Gildong was hiking a mountain, walking by millions of trees. Inspired by them, he suddenly came up with an interesting idea for trees in data structures: What if we add another edge in a tree?

Then he found that such tree-like graphs are called 1-trees. Since Gildong was bored of solving too many tree problems, he wanted to see if similar techniques in trees can be used in 1-trees as well. Instead of solving it by himself, he's going to test you by providing queries on 1-trees.

First, he'll provide you a tree (not 1-tree) with n vertices, then he will ask you q queries. Each query contains 5 integers: x, y, a, b, and k. This means you're asked to determine if there exists a path from vertex a to b that contains exactly k edges after adding a bidirectional edge between vertices x and y. A path can contain the same vertices and same edges multiple times. All queries are independent of each other; i.e. the added edge in a query is removed in the next query.

Input
The first line contains an integer n ($3 \le n \le 10^5$), the number of vertices of the tree.

Next n−1 lines contain two integers u and v ($1 \leq u, v \leq n, u \neq v$) each, which means there is an edge between vertex u and v. All edges are bidirectional and distinct.

Next line contains an integer q ($1 \le q \le 10^5$), the number of queries Gildong wants to ask.

Next q lines contain five integers x, y, a, b, and k each ($1 \leq x, y, a, b \leq n, x \neq y, 1 \leq k \leq 10^{9}$) – the integers explained in the description. It is guaranteed that the edge between x and y does not exist in the original tree.

Output
For each query, print "YES" if there exists a path that contains exactly k edges from vertex a to b after adding an edge between vertices x and y. Otherwise, print "NO".

You can print each letter in any case (upper or lower).

Example

input
5
1 2
2 3
3 4
4 5
5
1 3 1 2 2
1 4 1 3 2
1 4 1 3 3
4 2 3 3 9
5 2 3 3 9
output
YES
YES
NO
YES
NO
Note
The image below describes the tree (circles and solid lines) and the added edges for each query (dotted lines).

Possible paths for the queries with "YES" answers are:

• 1-st query: 1 – 3 – 2
• 2-nd query: 1 – 2 – 3
• 4-th query: 3 – 4 – 2 – 3 – 4 – 2 – 3 – 4 – 2 – 3

1. a->b
2. a->x->y->b
3. a->y->x->b

## Code

#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e5 + 10;
struct node {
int to, nxt;
} v[maxn << 1];
int head[maxn], cnt = 1, n, k, q, lg[maxn], dep[maxn], f[maxn][25];
void add(int from, int to)
{
v[cnt].to = to;
v[cnt].nxt = head[from];
head[from] = cnt++;
}
void dfs(int x, int fa)
{
f[x][0] = fa;
dep[x] = dep[fa] + 1;
for (int i = 1; i <= lg[dep[x]]; i++) f[x][i] = f[f[x][i - 1]][i - 1];
for (int i = head[x]; i; i = v[i].nxt)
if (v[i].to != fa) dfs(v[i].to, x);
}
int lca(int x, int y)
{
if (dep[x] < dep[y]) swap(x, y);
while (dep[x] > dep[y]) x = f[x][lg[dep[x] - dep[y]] - 1];
if (x == y) return x;
for (int i = lg[dep[x]] - 1; i >= 0; i--)
if (f[x][i] != f[y][i])
x = f[x][i], y = f[y][i];
return f[x][0];
}
int dis(int x, int y)
{
return dep[x] + dep[y] - dep[lca(x, y)] * 2;
}
bool check(int x)
{
if (x <= k && ((x & 1) == (k & 1))) return 1;
return 0;
}
int main()
{
ios::sync_with_stdio(0);
cin >> n;
for (int i = 1, u, v; i < n; i++) {
cin >> u >> v;
add(u, v); add(v, u);
}
for (int i = 1; i <= n; i++)
lg[i] = lg[i - 1] + (1 << lg[i - 1] == i);
dfs(1, 0);
cin >> q;
while (q--) {
int a, b, x, y;
cin >> x >> y >> a >> b >> k;
if (check(dis(a, b)) || check(dis(a, x) + dis(y, b) + 1) || check(dis(a, y) + dis(x, b) + 1)) cout << "YES\n";
else cout << "NO\n";
}
return 0;
}


## LCA模板

#include <bits/stdc++.h>
using namespace std;
const int maxn = 5e5 + 10;
int n, m, s;
struct node {
int to, next;
} v[maxn << 1];
int head[maxn], cnt = 1, dep[maxn], fa[maxn][30], lg[maxn];
void add(int from, int to)
{
v[cnt].to = to;
v[cnt].next = head[from];
head[from] = cnt++;
}
void dfs(int x, int f)
{
fa[x][0] = f; dep[x] = dep[f] + 1;
for (int i = 1; i <= lg[dep[x]]; i++) fa[x][i] = fa[fa[x][i - 1]][i - 1];
for (int i = head[x]; i; i = v[i].next) if (v[i].to != f) dfs(v[i].to, x);
}
int lca(int x, int y)
{
if (dep[x] < dep[y]) swap(x, y);
while (dep[x] > dep[y])
x = fa[x][lg[dep[x] - dep[y]] - 1];
if (x == y) return x;
for (int i = lg[dep[x]] - 1; i >= 0; i--)
if (fa[x][i] != fa[y][i])
x = fa[x][i], y = fa[y][i];
return fa[x][0];
}
int main()
{
ios::sync_with_stdio(0);
cin >> n >> m >> s;
for (int x, y, i = 1; i <= n - 1; i++) {
cin >> x >> y;
add(x, y); add(y, x);
}
for (int i = 1; i <= n; i++) lg[i] = lg[i - 1] + (1 << lg[i - 1] == i);
dfs(s, 0);
for (int i = 1, x, y; i <= m; i++) {
cin >> x >> y;
cout << lca(x, y) << endl;
}
}