【BZOJ3196】【TYVJ1730】二逼平衡树

【题目链接】

• 点击打开链接
  

【思路要点】

• 用外层树状数组套内层权值线段树，那么每一个区间对应的权值线段树可以表示为\(O(LogN)\)棵线段树的和（差）。
• 那么在得到的线段树上二分即可在\(O(LogNLogV)\)的时间内解决题目中所给出的询问。
• 注意修改仅涉及单点修改，那么至多会修改\(O(LogN)\)棵线段树，即修改的时间复杂度为\(O(LogNLogV)\)。
• 总时间复杂度\(O(MLogNLogV)\)，具体实现时需要对线段树进行动态开节点。

【代码】

#include<bits/stdc++.h>
using namespace std;
#define MAXN	50005
#define NODES	10000005
#define MAXV	100000005
struct Node {int lc,rc,sum;};
int n, m, size;
int value[MAXN], root[MAXN];
Node a[NODES];
int divide(int x, int y) {
	x += MAXV; y += MAXV;
	return (x + y) / 2 - MAXV;
}
int query(int root, int l, int r, int ql, int qr) {
	if (root == 0) return 0;
	if (l == ql && r == qr) return a[root].sum;
	int mid = divide(l, r), ans = 0;
	if (mid >= ql) ans += query(a[root].lc, l, mid, ql, min(qr, mid));
	if (mid + 1 <= qr) ans += query(a[root].rc, mid + 1, r, max(mid + 1, ql), qr);
	return ans;
}
int sum(int l, int r, int ql, int qr) {
	int ans = 0;
	for (int i = r; i >= 1; i -= i & -i)
		ans += query(root[i], -MAXV, MAXV, ql, qr);
	for (int i = l - 1; i >= 1; i -= i & -i)
		ans -= query(root[i], -MAXV, MAXV, ql, qr);
	return ans;
}
int find(int ql, int qr, int x) {
	static int v[MAXN], f[MAXN];
	int total = 0;
	for (int i = qr; i >= 1; i -= i & -i) {
		total++;
		v[total] = root[i];
		f[total] = 1;
	}
	for (int i = ql - 1; i >= 1; i -= i & -i) {
		total++;
		v[total] = root[i];
		f[total] = -1;
	}
	int l = -MAXV, r = MAXV;
	while (l < r) {
		int mid = divide(l, r), sum = 0;
		for (int i = 1; i <= total; i++)
			sum += f[i] * a[a[v[i]].lc].sum;
		if (x <= sum) {
			r = mid;
			for (int i = 1; i <= total; i++)
				v[i] = a[v[i]].lc;
		} else {
			x -= sum; l = mid + 1;
			for (int i = 1; i <= total; i++)
				v[i] = a[v[i]].rc;
		}
	}
	return l;
}
void maintain(int &root, int l, int r, int pos, int d) {
	if (root == 0) root = ++size;
	a[root].sum += d;
	if (l == r) return;
	int mid = divide(l, r);
	if (mid >= pos) maintain(a[root].lc, l, mid, pos, d);
	else maintain(a[root].rc, mid + 1, r, pos, d);
}
void maintain(int pos, int value, int d) {
	for (int i = pos; i <= n; i += i & -i)
		maintain(root[i], -MAXV, MAXV, value, d);
}
int main() {
	scanf("%d%d", &n, &m);
	for (int i = 1; i <= n; i++) {
		scanf("%d", &value[i]);
		maintain(i, value[i], 1);
	}
	for (int i = 1; i <= m; i++) {
		int opt, l, r, x, pos;
		scanf("%d", &opt);
		if (opt == 1) {
			scanf("%d%d%d", &l, &r, &x);
			printf("%d\n", sum(l, r, -MAXV, x - 1) + 1);
		}
		if (opt == 2) {
			scanf("%d%d%d", &l, &r, &x);
			printf("%d\n", find(l, r, x));
		}
		if (opt == 3) {
			scanf("%d%d", &pos, &x);
			maintain(pos, value[pos], -1);
			value[pos] = x;
			maintain(pos, value[pos], 1);
		}
		if (opt == 4) {
			scanf("%d%d%d", &l, &r, &x);
			printf("%d\n", find(l, r, sum(l, r, -MAXV, x - 1)));
		}
		if (opt == 5) {
			scanf("%d%d%d", &l, &r, &x);
			printf("%d\n",find(l, r, sum(l, r, -MAXV, x) + 1));
		}
	}
	return 0;
}