题意：

给\(n(1 \leq n \leq 10^5)\)一棵树，每个点有个权值。

还有\(m(1 \leq m \leq 10^5)\)个询问：

\(u \, v \, k\)，查询路径\(u \to v\)上权值第\(k\)小的权值。

分析：

和普通的区间一样，对于树维护到根节点的路径信息，父亲节点代表的树就是它的前一棵树。

查询的时候还要求出\(LCA(u,v)\)，路径上的点数就是：

• \(u\)到根的个数+\(v\)到根的个数-\(lca\)到根的个数\(\times 2\)

如果\(lca\)的权值也在统计范围内，统计个数再加\(1\)。

#include 
    
     #include 
     
      #include 
      
       using namespace std;const int maxn = 100000 + 10;struct Edge{    int v, nxt;    Edge() {}    Edge(int v, int nxt): v(v), nxt(nxt) {}};int head[maxn], ecnt;Edge edges[maxn * 2];void AddEdge(int u, int v) {    edges[ecnt] = Edge(v, head[u]);    head[u] = ecnt++;}int fa[maxn], L[maxn];void dfs(int u) {    for(int i = head[u]; ~i; i = edges[i].nxt) {        int v = edges[i].v;        if(v == fa[u]) continue;        fa[v] = u;        L[v] = L[u] + 1;        dfs(v);    }}int n, m;int anc[maxn][20];void preprocess() {    memset(anc, 0, sizeof(anc));    for(int i = 1; i <= n; i++) anc[i][0] = fa[i];    for(int j = 1; (1 << j) < n; j++)        for(int i = 1; i <= n; i++) if(anc[i][j-1])            anc[i][j] = anc[anc[i][j-1]][j-1];}int LCA(int u, int v) {    if(L[u] < L[v]) swap(u, v);    int log;    for(log = 0; (1 << log) < L[u]; log++);    for(int i = log; i >= 0; i--)        if(L[u] - (1<
       
        = L[v]) u = anc[u][i];    if(u == v) return u;    for(int i = log; i >= 0; i--)        if(anc[u][i] && anc[u][i] != anc[v][i])            u = anc[u][i], v = anc[v][i];    return fa[u];}int a[maxn], b[maxn], tot;struct Node{    int lch, rch, sum;};int sz;Node T[maxn << 5];int root[maxn];int newnode() {    sz++;    T[sz].lch = T[sz].rch = T[sz].sum = 0;    return sz;}int update(int pre, int L, int R, int p) {    int rt = newnode();    T[rt].lch = T[pre].lch;    T[rt].rch = T[pre].rch;    T[rt].sum = T[pre].sum + 1;    if(L == R) return rt;    int M = (L + R) / 2;    if(p <= M) T[rt].lch = update(T[pre].lch, L, M, p);    else T[rt].rch = update(T[pre].rch, M+1, R, p);    return rt;}void dfs2(int u) {    root[u] = update(root[fa[u]], 1, tot, a[u]);    for(int i = head[u]; ~i; i = edges[i].nxt) {        int v = edges[i].v;        if(v == fa[u]) continue;        dfs2(v);    }}int tmp;int query(int u, int v, int l, int L, int R, int k) {    if(L == R) return L;    int M = (L + R) / 2;    int sum = T[T[u].lch].sum + T[T[v].lch].sum - T[T[l].lch].sum * 2;    if(L <= tmp && tmp <= M) sum++;    if(sum >= k) return query(T[u].lch, T[v].lch, T[l].lch, L, M, k);    else return query(T[u].rch, T[v].rch, T[l].rch, M+1, R, k - sum);}int main(){    scanf("%d%d", &n, &m);    for(int i = 1; i <= n; i++) {        scanf("%d", a + i);        b[i] = a[i];    }    ecnt = 0;    memset(head, -1, sizeof(head));    for(int i = 1; i < n; i++) {        int u, v; scanf("%d%d", &u, &v);        AddEdge(u, v); AddEdge(v, u);    }    dfs(1);    preprocess();    sort(b + 1, b + 1 + n);    tot = unique(b + 1, b + 1 + n) - b - 1;    for(int i = 1; i <= n; i++)        a[i] = lower_bound(b + 1, b + 1 + tot, a[i]) - b;    dfs2(1);        while(m--) {        int u, v, k; scanf("%d%d%d", &u, &v, &k);        int lca = LCA(u, v);        tmp = a[lca];        int ans = query(root[u], root[v], root[lca], 1, tot, k);        printf("%d\n", b[ans]);    }    return 0;}