[BZOJ2733] [HNOI2012]永无乡（并查集 + 线段树合并）

传送门

一看到第k大就肯定要想到什么权值线段树，主席树，平衡树之类的

然后就简单了

用并查集判断连通，每个节点建立一颗权值线段树，连通的时候直接合并即可

查询时再二分递归地查找

时间复杂度好像不是很稳定。。。但hzwer都用这种方法水过。。

正解好像是平衡树+启发式合并，以后学TT

#include <cstdio>
#include <iostream>
#define N 100001

int n, m, q, cnt;
int a[N], f[N], sum[N * 20], ls[N * 20], rs[N * 20], root[N], id[N];

inline int read()
{
	int x = 0, f = 1;
	char ch = getchar();
	for(; !isdigit(ch); ch = getchar()) if(ch == '-') f = -1;
	for(; isdigit(ch); ch = getchar()) x = (x << 1) + (x << 3) + ch - '0';
	return x * f;
}

inline void insert(int &now, int l, int r, int x)
{
	now = ++cnt;
	if(l == r)
	{
		sum[now] = 1;
		return;
	}
	int mid = (l + r) >> 1;
	if(x <= mid) insert(ls[now], l, mid, x);
	else insert(rs[now], mid + 1, r, x);
	sum[now] = sum[ls[now]] + sum[rs[now]];
}

inline void merge(int &x, int y)
{
	if(!x || !y)
	{
		x += y;
		return;
	}
	sum[x] += sum[y];
	merge(ls[x], ls[y]);
	merge(rs[x], rs[y]);
}

inline int query(int now, int l, int r, int x)
{
	if(l == r) return l;
	int mid = (l + r) >> 1;
	if(sum[ls[now]] >= x)
		return query(ls[now], l, mid, x);
	else
		return query(rs[now], mid + 1, r, x - sum[ls[now]]);
}

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

int main()
{
	int i, x, y;
	char s[1];
	n = read();
	m = read();
	for(i = 1; i <= n; i++)
	{
		f[i] = i;
		x = read();
		id[x] = i;
		insert(root[i], 1, n, x);
	}
	for(i = 1; i <= m; i++)
	{
		x = find(read());
		y = find(read());
		if(x ^ y)
		{
			f[y] = x;
			merge(root[x], root[y]);
		}
	}
	q = read();
	while(q--)
	{
		scanf("%s", s);
		x = read();
		y = read();
		if(s[0] == 'B')
		{
			x = find(x);
			y = find(y);
			if(x ^ y)
			{
				f[y] = x;
				merge(root[x], root[y]);
			}
		}
		else
		{
			x = find(x);
			if(y > sum[root[x]]) puts("-1");
			else printf("%d\n", id[query(root[x], 1, n, y)]);
		}
	}
	return 0;
}