题面大意

Solution

#pragma GCC target("avx,sse2,sse3,sse4,popcnt")
#pragma GCC optimize("O2,O3,Ofast,inline,unroll-all-loops,-ffast-math")
#include <bits/stdc++.h>
using namespace std;
#define js ios::sync_with_stdio(false);cin.tie(0); cout.tie(0)
#define all(__vv__) (__vv__).begin(), (__vv__).end()
#define endl "\n"
#define pai pair<int, int>
#define mk(__x__,__y__) make_pair(__x__,__y__)
#define ms(__x__,__val__) memset(__x__, __val__, sizeof(__x__))
typedef long long ll; typedef unsigned long long ull; typedef long double ld;
inline ll read() { ll s = 0, w = 1; char ch = getchar(); for (; !isdigit(ch); ch = getchar()) if (ch == '-') w = -1; for (; isdigit(ch); ch = getchar())    s = (s << 1) + (s << 3) + (ch ^ 48); return s * w; }
inline void print(ll x, int op = 10) { if (!x) { putchar('0'); if (op)    putchar(op); return; }    char F[40]; ll tmp = x > 0 ? x : -x;    if (x < 0)putchar('-');    int cnt = 0;    while (tmp > 0) { F[cnt++] = tmp % 10 + '0';        tmp /= 10; }    while (cnt > 0)putchar(F[--cnt]);    if (op)    putchar(op); }
inline ll gcd(ll x, ll y) { return y ? gcd(y, x % y) : x; }
ll qpow(ll a, ll b) { ll ans = 1;    while (b) { if (b & 1)    ans *= a;        b >>= 1;        a *= a; }    return ans; }    ll qpow(ll a, ll b, ll mod) { ll ans = 1; while (b) { if (b & 1)(ans *= a) %= mod; b >>= 1; (a *= a) %= mod; }return ans % mod; }
inline int lowbit(int x) { return x & (-x); }
const int dir[][2] = { {0,1},{1,0},{0,-1},{-1,0},{1,1},{1,-1},{-1,1},{-1,-1} };
const int MOD = 1e9 + 7;
const int INF = 0x3f3f3f3f;

const int N = 3e5 + 7;

vector<int> p[N];
int deep[N];
int fa[N][20]; //log2(100000)  = 16.

void dfs(int x, int y) { //dfs求到深度以及预处理x的倍增数组
    fa[x][0] = y;
    deep[x] = deep[y] + 1;
    for (int i = 1; i < 20; ++i)
        fa[x][i] = fa[fa[x][i - 1]][i - 1];
    for (int i = 0; i < p[x].size(); ++i) {
        int tmp = p[x][i];
        if (tmp != y)
            dfs(tmp, x);
    }
}

int lca(int x, int y) {
    if (deep[x] < deep[y])   swap(x, y);
    for (int i = 19; i >= 0; --i)
        if (deep[fa[x][i]] >= deep[y])
            x = fa[x][i]; //调整到同一深度
    if (x == y) return x;
    for (int i = 19; i >= 0; --i)
        if (fa[x][i] != fa[y][i])
            x = fa[x][i], y = fa[y][i];
    return fa[x][0];
}

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

int main() {
    int n = read();
    for (int i = 1; i < n; ++i) {
        int u = read(), v = read();
        p[u].push_back(v);
        p[v].push_back(u);
    }
    dfs(1, 0);
    int u = read(), v = read();
    int T = read();
    while (T--) {
        int x = read(), y = read();
        int ans = dis(x, y);
        ans = min(ans, dis(x, u) + dis(y, v));
        ans = min(ans, dis(x, v) + dis(y, u));
        print(ans);
    }
    return 0;
}