[洛谷P3224][HNOI2012]永无乡

题目大意：给你$n$个点，每个点有权值$k$，现有两种操作：

1. $B;x;y：$将$x,y$所在联通块合并
2. $Q;x;k：$查询第$x$个点所在联通块权值第$k$小是哪个数

题解：线段树合并，权值线段树上二分即可

卡点：无

C++ Code：

#include <cstdio>
#include <cctype>
namespace __IO {
	namespace R {
		int x, ch;
		inline int read() {
			ch = getchar();
			while (isspace(ch)) ch = getchar();
			for (x = ch & 15, ch = getchar(); isdigit(ch); ch = getchar()) x = x * 10 + (ch & 15);
			return x;
		}
		inline char readc() {
			ch = getchar();
			while (!isalpha(ch)) ch = getchar();
			return ch;
		}
	}
}
using __IO::R::read;
using __IO::R::readc;

#define maxn 100010
#define N (maxn * 50)

int n, m;
int w[maxn], ret[maxn];
int rt[maxn], lc[N], rc[N], sz[N], idx;

int f[maxn];
int find(int x) {return x == f[x] ? x : (f[x] = find(f[x]));}

void insert(int &rt, int l, int r, int num) {
	if (!rt) rt = ++idx;
	sz[rt]++;
	if (l == r) return ;
	int mid = l + r >> 1;
	if (num <= mid) insert(lc[rt], l, mid, num);
	else insert(rc[rt], mid + 1, r, num);
}

int __merge(int x, int y) {
	if (!x || !y) return x | y;
	sz[x] += sz[y];
	lc[x] = __merge(lc[x], lc[y]);
	rc[x] = __merge(rc[x], rc[y]);
	return x;
}
void merge(int a, int b) {
	int x = find(a), y = find(b);
	if (x == y) return ;
	rt[x] = __merge(rt[x], rt[y]);
	f[y] = x;
}
int __query(int rt, int l, int r, int k) {
	if (sz[rt] < k) return -1;
	if (l == r) return l;
	int mid = l + r >> 1;
	if (sz[lc[rt]] >= k) return __query(lc[rt], l, mid, k);
	else return __query(rc[rt], mid + 1, r, k - sz[lc[rt]]);
}
int query(int x, int k) {
	x = find(x);
	int res = __query(rt[x], 1, n, k);
	return ~res ? ret[res] : -1;
}

int main() {
	n = read(), m = read();
	for (int i = 1; i <= n; i++) {
		w[i] = read(), ret[w[i]] = f[i] = i;
		insert(rt[i], 1, n, w[i]);
	}
	for (int i = 0, a, b; i < m; i++) {
		a = read(), b = read();
		merge(a, b);
	}
	int q = read();
	while (q --> 0) {
		char op = readc();
		int x = read(), y = read();
		if (op == 'B') merge(x, y);
		else printf("%d
", query(x, y));
	}
	return 0;
}