P4175 [CTSC2008]网络管理

动态链上第 $k$ 大。考虑树状数组 $+$ 主席树。

树状数组上每一个节点都是一棵主席树。

对于修改，树状数组上区间修改即可，$\mathcal O(\log^2 n)$。

对于查询，差分，主席树求第 $k$ 大，$\mathcal O(\log^2 n)$。

时间复杂度 $\mathcal O(n \log^2 n)$。

#include <bits/stdc++.h>

using namespace std;

namespace Fread {
	const int SIZE = 1 << 23;
	char buf[SIZE], *S, *T;
	inline char getchar() {
		if (S == T) {
			T = (S = buf) + fread(buf, 1, SIZE, stdin);
			if (S == T)
				return '\n';
		}
		return *S++;
	}
}
namespace Fwrite {
	const int SIZE = 1 << 23;
	char buf[SIZE], *S = buf, *T = buf + SIZE;
	inline void flush() {
		fwrite(buf, 1, S - buf, stdout);
		S = buf;
	}
	inline void putchar(char c) {
		*S++ = c;
		if (S == T)
			flush();
	}
	struct NTR {
		~NTR() {
			flush();
		}
	} ztr;
}

#ifdef ONLINE_JUDGE
	#define getchar Fread::getchar
	#define putchar Fwrite::putchar
#endif

inline int read() {
	int x = 0, f = 1;
	char c = getchar();
	while (c < '0' || c > '9') {
		if (c == '-')
			f = -1;
		c = getchar();
	}
	while (c >= '0' && c <= '9')
		x = x * 10 + c - '0', c = getchar();
	return x * f;
}

inline void write(int x) {
	if (x < 0) {
		putchar('-');
		x = -x;
	}
	if (x > 9)
		write(x / 10);
	putchar(x % 10 + '0');
}

const int _ = 1e5 + 10;

int n, m, cnt, len, a[_], b[_];

int tot, head[_], to[_ << 1], nxt[_ << 1];

int lc[_ * 200], rc[_ * 200], sum[_ * 200];

int rt[_], dep[_], fa[_][20], dfc, st[_], ed[_];

int c[2], tmp[2][_];

struct que {
	int k, a, b;
} q[_];

inline void add(int u, int v) {
	to[++tot] = v, nxt[tot] = head[u], head[u] = tot;
	to[++tot] = u, nxt[tot] = head[v], head[v] = tot;
}

void update(int &u, int l, int r, int pos, int val) {
	if (!u) u = ++cnt;
	sum[u] += val;
	if (l == r) return;
	int mid = (l + r) >> 1;
	if (pos <= mid) update(lc[u], l, mid, pos, val);
	else update(rc[u], mid + 1, r, pos, val);
}

int query(int l, int r, int k) {
	if (l == r) return l;
	int mid = (l + r) >> 1, s = 0;
	for (int i = 1; i <= c[0]; ++i) s += sum[lc[tmp[0][i]]];
	for (int i = 1; i <= c[1]; ++i) s -= sum[lc[tmp[1][i]]];
	if (k <= s) {
		for (int i = 1; i <= c[0]; ++i) tmp[0][i] = lc[tmp[0][i]];
		for (int i = 1; i <= c[1]; ++i) tmp[1][i] = lc[tmp[1][i]];
		return query(l, mid, k);
	} else {
		for (int i = 1; i <= c[0]; ++i) tmp[0][i] = rc[tmp[0][i]];
		for (int i = 1; i <= c[1]; ++i) tmp[1][i] = rc[tmp[1][i]];
		return query(mid + 1, r, k - s);
	}
}

void dfs(int x) {
	for (int i = 1; i < 20; ++i) fa[x][i] = fa[fa[x][i - 1]][i - 1];
	dep[x] = dep[fa[x][0]] + 1, st[x] = ++dfc;
	for (int i = head[x]; i; i = nxt[i]) {
		int v = to[i];
		if (v == fa[x][0]) continue;
		fa[v][0] = x;
		dfs(v);
	}
	ed[x] = dfc;
}

int LCA(int x, int y) {
	if (dep[x] < dep[y]) swap(x, y);
	int t = dep[x] - dep[y];
	for (int i = 0; i < 20; ++i)
		if (t & (1 << i)) x = fa[x][i];
	if (x == y) return x;
	for (int i = 19; ~i; --i)
		if (fa[x][i] != fa[y][i]) x = fa[x][i], y = fa[y][i];
	return fa[x][0];
}

inline int lowbit(int x) {
	return x & (-x);
}

inline void change(int pos, int val, int op) {
	for (; pos <= n; pos += lowbit(pos)) update(rt[pos], 1, len, val, op);
}

signed main() {
	n = read(), m = read(), len = n;
	for (int i = 1; i <= n; ++i) a[i] = b[i] = read();
	for (int i = 1; i < n; ++i) add(read(), read());
	for (int i = 1; i <= m; ++i) {
		q[i].k = read(), q[i].a = read(), q[i].b = read();
		if (!q[i].k) b[++len] = q[i].b;
	}
	sort(b + 1, b + len + 1), len = unique(b + 1, b + len + 1) - b - 1;
	for (int i = 1; i <= n; ++i) a[i] = lower_bound(b + 1, b + len + 1, a[i]) - b;
	dfs(1);
	for (int i = 1; i <= n; ++i) change(st[i], a[i], 1), change(ed[i] + 1, a[i], -1);
	for (int i = 1; i <= m; ++i) {
		int x = q[i].a, y = q[i].b, k = q[i].k, lca, flca, s;
		if (k) {
			lca = LCA(x, y), flca = fa[lca][0], c[0] = c[1] = 0;
			s = dep[x] + dep[y] - dep[lca] - dep[flca];
			if (k > s) {
				printf("invalid request!\n");
				continue;
			}
			for (int i = st[x]; i; i -= lowbit(i)) tmp[0][++c[0]] = rt[i];
			for (int i = st[y]; i; i -= lowbit(i)) tmp[0][++c[0]] = rt[i];
			for (int i = st[lca]; i; i -= lowbit(i)) tmp[1][++c[1]] = rt[i];
			for (int i = st[flca]; i; i -= lowbit(i)) tmp[1][++c[1]] = rt[i];
			printf("%d\n", b[query(1, len, s - k + 1)]);
		} else {
			change(st[x], a[x], -1), change(ed[x] + 1, a[x], 1);
			a[x] = lower_bound(b + 1, b + len + 1, y) - b;
			change(st[x], a[x], 1), change(ed[x] + 1, a[x], -1);
		}
	}
	return 0;
}