題意：給出一棵樹，然後給出兩個數對 $(a, b)$, $(c$

題解：才開始想要標記路徑但是LCA是倍增的不好弄，時間複雜度也太高，然後想了一下樹上求最短路的知識，然後想到他們的LCA的高度，然後通過交叉走判斷是否有重合的點。具體先選擇LCA較小的點作為比較物件，然後另外兩個點向當前兩個點求LCA看高度與最低的LCA路徑和，最後別忘了加上1（祖先）

$ac code:$

#include
 
 <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define met(a, b) memset(a, b, sizeof(a))
#define rep(i, a, b) for(int i = a; i <= b; i++)
#define per(i, a, b) for(int i = a; i >= b; i--)
#define debug cout << "*" << endl;

const int maxn = 1e5 + 10;
const int inf = 0x3f3f3f3f;
const int DEG = 20;

struct Edge {
    int to, next;
} edge[maxn << 1];

int head[maxn], tot;
void addEdge(int u, int v) {
    edge[tot].to = v;
    edge[tot].next = head[u];
    head[u] = tot++;
}

void init() {
    tot = 0;
    met(head, -1);
}

int fa[maxn][DEG], deg[maxn];

void BFS(int root) {
    queue<int> que;
    deg[root] = 0;
    fa[root][0] = root;
    que.push(root);
    while(!que.empty()) {
        int tmp = que.front();
        que.pop();
        for(int i = 1; i < DEG; i++) fa[tmp][i] = fa[fa[tmp][i - 1]][i - 1];
        for(int i = head[tmp]; i != -1; i = edge[i].next) {
            int v = edge[i].to;
            //cout << v << endl;
            if(v == fa[tmp][0]) continue;
            deg[v] = deg[tmp] + 1;
            fa[v][0] = tmp;
            que.push(v);
        }
    }
}

int LCA(int u, int v) {
    if(deg[u] > deg[v]) swap(u, v);
    int hu = deg[u], hv = deg[v];
    int tu = u, tv = v;
    for(int det = hv - hu, i = 0; det; det >>= 1, i++) {
        if(det & 1)
            tv = fa[tv][i];
    }
    if(tu == tv) return tu;
    for(int i = DEG - 1; i >= 0; i--) {
        if(fa[tu][i] == fa[tv][i]) {
            continue;
        }
        tu = fa[tu][i];
        tv = fa[tv][i];
    }
    return fa[tu][0];
}

bool flag[maxn];
int N, Q;

int dis(int a, int b) {
    int f = LCA(a, b);
    return deg[a] + deg[b] - 2 * deg[f];
}

int solve(int a, int b, int c, int d) {
    int lca = LCA(c, d);
    int t = LCA(a, b);
    if(deg[lca] < deg[t]) {
        lca = t;
        swap(a, c);
        swap(b, d);
    }
    t = LCA(a, c);
    int res = 0;
    int f = 0;
    if(deg[t] > deg[lca]) {
        res += dis(t, lca);
    } else if(deg[t] == deg[lca]) {
        f = 1;
    }
    t = LCA(a, d);
    if(deg[t] > deg[lca]) {
        res += dis(t, lca);
    } else if(deg[t] == deg[lca]) {
        f = 1;
    }
    t = LCA(b, c);
    if(deg[t] > deg[lca]) {
        res += dis(t, lca);
    } else if(deg[t] == deg[lca]) {
        f = 1;
    }
    t = LCA(b, d);
    if(deg[t] > deg[lca]) {
        res += dis(t, lca);
    } else if(deg[t] == deg[lca]) {
        f = 1;
    }
    res += f;
    return res;
}
int main() {
    int u, v, a, b, c, d;
    while(~scanf("%d%d", &N, &Q)) {
        init();
        met(flag, false);
        rep(i, 2, N) scanf("%d%d", &u, &v), addEdge(u, v), addEdge(v, u), flag[v] = true;
        int root;
        for(int i = 1; i <= N; i++) {
            if(!flag[i]) {
                root = i;
                break;
            }
        }
        BFS(root);
        while(Q--) {
            scanf("%d%d%d%d", &a, &b, &c, &d);
            printf("%d\n", solve(a, b, c, d));
        }
    }
    return 0;
}