HNOI2012 永无乡

题意

给定(n)个连通块，有两种操作：

• 合并两个连通块
• 查询某个元素所在连通块内第(k)大的值

解法

合并连通块( o)启发式合并，查询第(k)大( o)平衡树，权值线段树

当然这道题可以用线段树合并写，但是用FHQ_Treap来写实在是太爽了

由于FHQ_Treap本身就可以维护连通块（一颗树就是一个连通块），还能顺带维护连通块的size与其中的元素，简直是为这道题量身定制的

对于合并两个连通块的操作，我们遍历较小的那个连通块，将其中的元素一个个加进大连通块中，并把合并以后得到的根在并查集中设为原来连通块根的祖先

代码

#include <ctime>
#include <cstdio>
#include <cstdlib>
#include <iostream>

using namespace std;

const int N = 1e5 + 10;

int read();

int n, m;

int fa[N], mp[N];

char op[10];

struct FHQ_Treap {
#define ls(x) t[x].ch[0]
#define rs(x) t[x].ch[1]

	int cnt;
	int a, b;
	
	struct node {
		int val, rnd, siz;
		int ch[2];
		node() { ch[0] = ch[1] = 0; }	
	} t[N];
	
	FHQ_Treap() : cnt(0) {}
	
	void update(int x) {
		t[x].siz = t[ls(x)].siz + t[rs(x)].siz + 1;	
	}
	
	int newnode(int v) {
		++cnt;
		t[cnt].val = v, t[cnt].rnd = rand() << 15 | rand(), t[cnt].siz = 1;
		return cnt; 
	}
	
	void split(int x, int k, int& lt, int& rt) {
		if (!x)
			return lt = rt = 0, void();
		if (t[x].val <= k)
			lt = x, split(rs(x), k, rs(x), rt);
		else
			rt = x, split(ls(x), k, lt, ls(x));
		update(x);
	}
	
	int merge(int x, int y) {
		if (!x || !y)
			return x | y;
		if (t[x].rnd < t[y].rnd) {
			rs(x) = merge(rs(x), y); 
			return update(x), x;
		} else {
			ls(y) = merge(x, ls(y));	
			return update(y), y;
		}
	}
	
	int kth(int x, int k) {
		if (k > t[x].siz)  return 0;
		int p = x;	
		while (true) {
			if (k <= t[ls(p)].siz)  p = ls(p);
			else if (k == t[ls(p)].siz + 1)  return t[p].val;
			else k -= t[ls(p)].siz + 1, p = rs(p);
		}
	}
	
	int insert(int x, int y) {
		split(y, t[x].val, a, b);
		return merge(merge(a, x), b);
	}
	
	void DFS(int x, int& y) {
		if (ls(x))  DFS(ls(x), y);
		if (rs(x))  DFS(rs(x), y);
		y = insert(x, y);
	}
	
	int combine(int x, int y) {
		DFS(x, y);
		return y;
	}
	
#undef ls
#undef rs	
} tr;

inline int get(int x) {
	return x == fa[x] ? x : fa[x] = get(fa[x]);	
}

void modify(int u, int v) {
	if (get(u) ^ get(v)) {
		u = fa[u], v = fa[v];
		if (tr.t[u].siz > tr.t[v].siz)  swap(u, v);
		int nr = tr.combine(u, v);
		fa[u] = fa[v] = fa[nr] = nr;
	}
}

int main() {
	
	srand(time(0));
	
	n = read(), m = read();	
	
	mp[0] = -1;
	for (int i = 1; i <= n; ++i) {
		int v = read();
		fa[i] = mp[v] = i;
		tr.newnode(v);
	}
	
	for (int i = 1; i <= m; ++i)  modify(read(), read());
	
	int q = read();
	while (q--) {
		scanf("%s", op + 1);
		if (op[1] == 'B') {
			modify(read(), read());
		} else {
			int u = read(), k = read();
			printf("%d
", mp[tr.kth(get(u), k)]);
		}
	}
	
	return 0;
}

int read() {
	int x = 0, c = getchar();
	while (c < '0' || c > '9')    c = getchar();
	while (c >= '0' && c <= '9')  x = x * 10 + c - 48, c = getchar();
	return x;	
}