前言：一开始看到这个题目的时候，感觉就和lca有关，但是没有想到具体的公式

$d=dep[x]+dep[y]-2\ast dep[lca(x,y)]$

并且我们这个题目还是一个进阶版本的题，我们还会存在电缆，我们还有两种选择的可能，这也是要考虑在内的

题目地址

#include <bits/stdc++.h>
using namespace std;
#define ll long long

const int N = (int)5e5;
int n;
int e[N * 2], ne[N * 2], h[N], idx = 0;
int px, py;
int dep[N], fa[N][20];

inline int read()
{
	int x = 0, f = 1;
	char c = getchar();
	while (c < '0' || c>'9') { if (c == '-') f = -1; c = getchar(); }
	while (c >= '0' && c <= '9') x = (x << 1) + (x << 3) + c - '0', c = getchar();
	return f * x;
}

void add(int a, int b) {
	e[++idx] = b, ne[idx] = h[a], h[a] = idx;
}

void dfs(int u, int father) {
	dep[u] = dep[father] + 1;
	fa[u][0] = father;
	for (int i = 1; i < 20; i++) {
		fa[u][i] = fa[fa[u][i - 1]][i - 1];
	}
	for (int i = h[u]; i; i = ne[i]) {
		int to = e[i];
		if (to == father) continue;
		dfs(to, u);
	}
}

int lca(int u, int v) {
	if (dep[u] < dep[v]) swap(u, v);
	// 先跳到同一层
	for (int i = 19; i >= 0; i--) {
		if (dep[fa[u][i]] >= dep[v]) u = fa[u][i];
	}
	if (u == v) return u;
	for (int i = 19; i >= 0; i--) {
		if (fa[u][i] != fa[v][i]) {
			u = fa[u][i], v = fa[v][i];
		}
	}return fa[u][0];
}

int fun(int x, int y) {
	return dep[x] + dep[y] - 2 * dep[lca(x, y)];
}

int main() {
	cin >> n;
	for (int i = 1; i < n; i++) {
		int x, y; 
		//cin >> x >> y;
		x = read(); y = read();
		add(x, y), add(y, x);
	}
	cin >> px >> py;
	int q; cin >> q;
	dfs(1, 0);
	while (q--) {
		int x, y; 
		//cin >> x >> y;
		x = read(); y = read();
		int d = min({ fun(x,y),fun(x,px) + fun(y,py),fun(x,py) + fun(y,px) });
		//cout << d << endl;
		printf("%d\n", d);
	}return 0;
}