小A的最短路

思路

树上问题求两个点的最短距离，显然能用$lca$来进行$lo{g}_n$的查询，引入了两个无边权的点，所以我们的路劲就可以规划成三种$x->y，x->u->v->y，x->v->u>-y$，只要在这三个当中取一个最小值就行了。接下来就是考虑求$lca$了，有一种较为快速的求$lca$的在线方法，那就是树链剖分，于是套上去(个人认为树剖求$lca$较为好写)，然后就可以开始最短路求解了。

代码

/*Author : lifehappy
*/
#pragma GCC optimize(2)
#pragma GCC optimize(3)
#include 
#define mp make_pair
#define pb push_back
#define endl '\n'using namespace std;typedef long long ll;
typedef unsigned long long ull;
typedef pair pii;const double pi = acos(-1.0);
const double eps = 1e-7;
const int inf = 0x3f3f3f3f;inline ll read() {ll f = 1, x = 0;char c = getchar();while(c  '9') {if(c == '-') f = -1;c = getchar();}while(c >= '0' && c <= '9') {x = (x << 1) + (x << 3) + (c ^ 48);c = getchar();}return f * x;
}void print(ll x) {if(x < 10) {putchar(x + 48);return ;}print(x / 10);putchar(x % 10 + 48);
}const int N = 3e5 + 10;int sz[N], son[N], fa[N], dep[N], top[N], n, m;int head[N], to[N << 1], nex[N << 1], cnt = 1;void add(int x, int y) {to[cnt] = y;nex[cnt] = head[x];head[x] = cnt++;
}void dfs1(int rt, int f) {fa[rt] = f, sz[rt] = 1;dep[rt] = dep[f] + 1;for(int i = head[rt]; i; i = nex[i]) {if(to[i] == f) continue;dfs1(to[i], rt);sz[rt] += sz[to[i]];if(!son[rt] || sz[to[i]] > sz[son[rt]]) son[rt] = to[i];}
}void dfs2(int rt, int tp) {top[rt] = tp;if(!son[rt]) return ;dfs2(son[rt], tp);for(int i = head[rt]; i; i = nex[i]) {if(to[i] == fa[rt] || to[i] == son[rt]) continue;dfs2(to[i], to[i]);}
}int lca(int x, int y) {while(top[x] != top[y]) {if(dep[top[x]] < dep[top[y]]) swap(x, y);x = fa[top[x]];}return dep[x] < dep[y] ? x : y;
}int dis(int x, int y) {return dep[x] + dep[y] - 2 * dep[lca(x, y)];
}int main() {// freopen("in.txt", "r", stdin);// freopen("out.txt", "w", stdout);//ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);n = read();for(int i = 1; i < n; i++) {int x = read(), y = read();add(x, y);add(y, x);}int u = read(), v = read();dfs1(1, 0);dfs2(1, 1);m = read();for(int i = 1; i <= m; i++) {int x = read(), y = read();printf("%d\n", min({dis(x, y), dis(x, u) + dis(v, y), dis(x, v) + dis(u, y)}));}return 0;
}