HNOI2012 永无乡

( ext{Problem})

[HNOI2012]永无乡

( ext{Solution})

学了线段树合并后发现它是个模板
它确实模板
于是我们只要维护若干棵权值线段树，询问直接找，连边合并就好
并查集维护一坨一坨的关系
注意并查集合并和线段树合并的方向一致即可

( ext{Code})

#include<cstdio>
using namespace std;

const int N = 1e5 + 5;
int n, m, q, size, fa[N], id[N], rt[N];
int sum[N * 32], ls[N * 32], rs[N * 32];

inline void pushup(int p){sum[p] = sum[ls[p]] + sum[rs[p]];}

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

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

inline void dsu_union(int x, int y)
{
	int u = find(x), v = find(y);
	if (u == v) return;
	fa[v] = fa[u];
	merge(rt[u], rt[v]);
	rt[v] = 0;
}

void update(int &p, int l, int r, int x)
{
	if (!p) p = ++size;
	if (l == r) return void(++sum[p]);
	int mid = (l + r) >> 1;
	if (x <= mid) update(ls[p], l, mid, x);
	else update(rs[p], mid + 1, r, x);
	pushup(p);
}

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

int main()
{
	scanf("%d%d", &n, &m);
	for(int i = 1, x; i <= n; i++) 
		fa[i] = i, scanf("%d", &x), id[x] = i, update(rt[i], 1, n, x);
	for(int i = 1, x, y; i <= m; i++) 
		scanf("%d%d", &x, &y), dsu_union(x, y);
	scanf("%d", &q);
	char op[5];
	for(int i = 1, x, y; i <= q; i++)
	{
		scanf("%s%d%d", op, &x, &y);
		if (op[0] == 'Q') printf("%d
", query(rt[find(x)], 1, n, y));
		else dsu_union(x, y);
	}
}