本题求一棵树上两点之间简单路径上第k小的点权，强制在线。函数式线段树（主席树）是一种可持久化数据结构。第i棵线段树保存前缀[1..i]的信息，在树上，前缀可扩展为结点i到根的简单路径。具体地，第i棵线段树中的位置x为a，意味着[1..i]中有a个元素具有性质x——按权值建线段树。这样建出来的线段树有两个性质：
1. 形态相同。
2. 可加减，比如第i棵线段树-第j棵线段树=[j+1..i]。
因此，我们可以在第(i-1)棵线段树的基础上建第i棵线段树。离散化后，两者至多有lg n个结点不同；那些相同的结点，复用即可。自始至终，我们并未修改结点，只是新建和复用——函数式的思想。
新建一棵树的代码如下，其中MAXD=lg MAXN向上取整：
int ptr = 1, root[MAXN+1];struct Node {int v, lc, rc;
} T[MAXN*(MAXD+1)];void build(int& y, int x, int a, int l, int r)
{T[y = ptr++] = T[x];++T[y].v;if (l == r)return;int m = (l+r)/2;if (a <= m)build(T[y].lc, T[x].lc, a, l, m);elsebuild(T[y].rc, T[x].rc, a, m+1, r);
}查询，像普通线段树一样二分。其中lca(u, v)返回u和v的最近公共祖先，anc[a][0]是a的父亲，j是u、v间简单路径上权在[l, r]范围内的点数：
int query(int u, int v, int k)
{int l = 1, r = top, a = lca(u, v), x = root[u], y = root[v], z = root[a], t = root[anc[a][0]];while (l < r) {int m = (l+r)/2, j = T[T[x].lc].v + T[T[y].lc].v - T[T[z].lc].v - T[T[t].lc].v;if (j >= k) {x = T[x].lc;y = T[y].lc;z = T[z].lc;t = T[t].lc;r = m;} else {x = T[x].rc;y = T[y].rc;z = T[z].rc;t = T[t].rc;l = m+1;k -= j;}}return l;
}所以还需要找lca，这里采用倍增算法。完整代码如下：
#include <cstdio>
#include <algorithm>
using namespace std;
const int MAXN = 100000, MAXD = 17;
int N, M, e_ptr = 1, top = 1, maxd = 1;
int w[MAXN+1], h1[MAXN+1], h[MAXN+1], fst[MAXN+1];struct Edge {int v, next;
} E[MAXN*2];inline void add_edge(int u, int v)
{E[e_ptr] = (Edge){v, fst[u]}; fst[u] = e_ptr++;E[e_ptr] = (Edge){u, fst[v]}; fst[v] = e_ptr++;
}inline int read()
{int x = 0;char ch = getchar();while (ch<'0' || ch>'9')ch = getchar();while (ch>='0' && ch<='9') {x = x*10 + ch - '0';ch = getchar();}return x;
}namespace work {int ptr = 1, anc[MAXN+1][MAXD+1], root[MAXN+1], depth[MAXN+1];struct Node {int v, lc, rc;} T[MAXN*(MAXD+1)];void build(int& y, int x, int a, int l, int r){T[y = ptr++] = T[x];++T[y].v;if (l == r)return;int m = (l+r)/2;if (a <= m)build(T[y].lc, T[x].lc, a, l, m);elsebuild(T[y].rc, T[x].rc, a, m+1, r);}void dfs(int u, int fa){for (int i = 1; i <= maxd; ++i)anc[u][i] = anc[anc[u][i-1]][i-1];build(root[u], root[fa], lower_bound(h+1, h+top+1, w[u])-h, 1, top);for (int i = fst[u]; i; i = E[i].next) {int v = E[i].v;if (v != fa) {anc[v][0] = u;depth[v] = depth[u]+1;dfs(v, u);}}}inline void swim(int& x, int h){for (int i = 0; h; ++i, h >>= 1)if (h & 1)x = anc[x][i];}int lca(int u, int v){if (depth[u] > depth[v])swap(u, v);swim(v, depth[v]-depth[u]);for (int i = maxd; i >= 0; --i)if (anc[u][i] != anc[v][i]) {u = anc[u][i];v = anc[v][i];}if (u != v)u = anc[u][0];return u;}int query(int u, int v, int k){int l = 1, r = top, a = lca(u, v), x = root[u], y = root[v], z = root[a], t = root[anc[a][0]];while (l < r) {int m = (l+r)/2, j = T[T[x].lc].v + T[T[y].lc].v - T[T[z].lc].v - T[T[t].lc].v;if (j >= k) {x = T[x].lc;y = T[y].lc;z = T[z].lc;t = T[t].lc;r = m;} else {x = T[x].rc;y = T[y].rc;z = T[z].rc;t = T[t].rc;l = m+1;k -= j;}}return l;}
}int main()
{N = read();M = read();while ((1<<maxd) < N)++maxd;for (int i = 1; i <= N; ++i)h1[i] = w[i] = read();sort(h1+1, h1+N+1);h[1] = h1[1];for (int i = 2; i <= N; ++i)if (h1[i] != h1[i-1])h[++top] = h1[i];for (int i = 1; i < N; ++i) {int u = read(), v = read();add_edge(u, v);}work::dfs(1, 0);int lastans = 0;for (int i = 1; i <= M; ++i) {int u = read(), v = read(), k = read();u ^= lastans;printf("%d", lastans = h[work::query(u, v, k)]);if (i != M)putchar('\n');}return 0;
}最后不要换行，否则PE。