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 <cstdio>
#include <cstring>
#include <algorithm>
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<<i) >= 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;
}
