【BZOJ2733】【HNOI2012】永無鄉

【題目連結】

• 點選打開連結

【思路要點】

• 線段樹合并裸題。
• 時間複雜度\(O(NLogN+QLogN)\)。

【代碼】

#include<bits/stdc++.h>
using namespace std;
const int MAXN = 200005;
const int MAXP = 10000005;
template <typename T> void chkmax(T &x, T y) {x = max(x, y); }
template <typename T> void chkmin(T &x, T y) {x = min(x, y); } 
template <typename T> void read(T &x) {
	x = 0; int f = 1;
	char c = getchar();
	for (; !isdigit(c); c = getchar()) if (c == '-') f = -f;
	for (; isdigit(c); c = getchar()) x = x * 10 + c - '0';
	x *= f;
}
template <typename T> void write(T x) {
	if (x < 0) x = -x, putchar('-');
	if (x > 9) write(x / 10);
	putchar(x % 10 + '0');
}
template <typename T> void writeln(T x) {
	write(x);
	puts("");
}
struct SegmentTree {
	struct Node {
		int lc, rc;
		int cnt;
	} a[MAXP];
	int n, size;
	void init(int x) {
		n = x;
		size = 0;
	}
	void modify(int &root, int l, int r, int val) {
		if (root == 0) root = ++size;
		a[root].cnt++;
		if (l == r) return;
		int mid = (l + r) / 2;
		if (mid >= val) modify(a[root].lc, l, mid, val);
		else modify(a[root].rc, mid + 1, r, val);
	}
	int newtree(int val) {
		int ans = 0;
		modify(ans, 1, n, val);
		return ans;
	}
	int merge(int x, int y) {
		if (x == 0 || y == 0) return x + y;
		a[x].cnt += a[y].cnt;
		a[x].lc = merge(a[x].lc, a[y].lc);
		a[x].rc = merge(a[x].rc, a[y].rc);
		return x;
	}
	int query(int root, int l, int r, int k) {
		if (l == r) return l;
		int mid = (l + r) / 2;
		if (k <= a[a[root].lc].cnt) return query(a[root].lc, l, mid, k);
		else return query(a[root].rc, mid + 1, r, k - a[a[root].lc].cnt);
	}
	int query(int root, int k) {
		if (a[root].cnt < k) return -1;
		else return query(root, 1, n, k);
	}
} ST;
int n, m, q, f[MAXN], root[MAXN], ans[MAXN];
int F(int x) {
	if (f[x] == x) return x;
	else return f[x] = F(f[x]);
}
int main() {
	read(n), read(m);
	ST.init(n);
	for (int i = 1; i <= n; i++) {
		int x; read(x);
		ans[x] = f[i] = i;
		root[i] = ST.newtree(x);
	}
	for (int i = 1; i <= m; i++) {
		int x, y; read(x), read(y);
		if (F(x) != F(y)) {
			root[F(y)] = ST.merge(root[F(x)], root[F(y)]);
			f[F(x)] = F(y);
		}
	}
	read(q);
	for (int i = 1; i <= q; i++) {
		char opt; int x, y;
		scanf("\n%c%d%d", &opt, &x, &y);
		if (opt == 'B') {
			if (F(x) != F(y)) {
				root[F(y)] = ST.merge(root[F(x)], root[F(y)]);
				f[F(x)] = F(y);
			}
		} else {
			int tmp = ST.query(root[F(x)], y);
			if (tmp == -1) writeln(tmp);
			else writeln(ans[tmp]);
		}
	}
	return 0;
}