洛谷 P3380 【模板】二逼平衡树（树套树）

题面

luogu

题解

2019年AC的第一道题~~
函数名命名为rank竟然会ce

我写的是树状数组套值域线段树(动态开点)

操作1:询问(k)在([l-r])这段区间有多少数比它小,再加(1)
操作2:前缀和思想得到([l-r])区间的线段树,然后类似平衡树找第(k)大
操作3:直接修改
操作4/5:操作1+操作2

Code

#include<bits/stdc++.h>

#define LL long long
#define RG register

const int N = 50010;

using namespace std;

inline int gi() {
	RG int x = 0; RG char c = getchar(); bool f = 0;
	while (c != '-' && (c < '0' || c > '9')) c = getchar();
	if (c == '-') c = getchar(), f = 1;
	while (c >= '0' && c <= '9') x = x*10+c-'0', c = getchar();
	return f ? -x : x;
}

int tot, A[N<<1], a[N], cnt, n;
int t1, t2, tmp1[N], tmp2[N];

struct question {
	int op, l, r, k;
}q[N];

struct node {
	int ls, rs, v;
}t[N<<7];
int rt[N];
#define lowbit(x) (x&(-x))
void Update(int &now, int l, int r, int pos, int k) {
	if (!now) now = ++cnt;
	t[now].v += k;
	if (l == r) return ;
	int mid = (l + r) >> 1;
	if (pos <= mid) Update(t[now].ls, l, mid, pos, k);
	else Update(t[now].rs, mid+1, r, pos, k);
}
void update(int x, int k) {
	for (int i = x; i <= n; i += lowbit(i)) Update(rt[i], 1, tot, a[x], k);
}
int query(int now, int l, int r, int pos) {
	if (l == r) return t[now].v;
	int mid = (l + r) >> 1;
	if (pos <= mid) return query(t[now].ls, l, mid, pos);
	return t[t[now].ls].v+query(t[now].rs, mid+1, r, pos);
}

int Rank(int l, int r, int k) {
	if (l > r) return 0;
	l--;
	int s = 0;
	for (int i = r; i; i -= lowbit(i)) s += query(rt[i], 1, tot, k);
	for (int i = l; i; i -= lowbit(i)) s -= query(rt[i], 1, tot, k);
	return s;
}

int Kth(int l, int r, int k) {
	if (l == r) return l;
	int mid = (l + r) >> 1, s = 0;
	for (int i = 1; i <= t1; i++) s += t[t[tmp1[i]].ls].v;
	for (int i = 1; i <= t2; i++) s -= t[t[tmp2[i]].ls].v;
	if (s >= k) {
		for (int i = 1; i <= t1; i++) tmp1[i] = t[tmp1[i]].ls;
		for (int i = 1; i <= t2; i++) tmp2[i] = t[tmp2[i]].ls;
		return Kth(l, mid, k);
	}
	else {
		for (int i = 1; i <= t1; i++) tmp1[i] = t[tmp1[i]].rs;
		for (int i = 1; i <= t2; i++) tmp2[i] = t[tmp2[i]].rs;
		return Kth(mid+1, r, k-s);
	}
}

int kth(int l, int r, int k) {
	l--; t1 = t2 = 0;
	for (int i = r; i; i -= lowbit(i)) tmp1[++t1] = rt[i];
	for (int i = l; i; i -= lowbit(i)) tmp2[++t2] = rt[i];
	return A[Kth(1, tot, k)];
}

int main() {
	//freopen(".in", "r", stdin);
	//freopen(".out", "w", stdout);
	n = gi();
	int m = gi();
	for (int i = 1; i <= n; i++) A[++tot] = a[i] = gi();
	for (int i = 1; i <= m; i++) {
		q[i].op = gi();
		if (q[i].op != 3) {
			q[i].l = gi(); q[i].r = gi(); q[i].k = gi();
			if (q[i].op != 2) A[++tot] = q[i].k;
		}
		else {
			q[i].l = q[i].r = gi();
			A[++tot] = q[i].k = gi();
		}
	}
	sort(A+1, A+1+tot);
	tot = unique(A+1, A+1+tot) - A - 1;
	for (int i = 1; i <= n; i++) a[i] = lower_bound(A+1, A+1+tot, a[i])-A;
	for (int i = 1; i <= n; i++) update(i, 1);
	for (int i = 1; i <= m; i++)
		if (q[i].op != 2)
			q[i].k = lower_bound(A+1, A+1+tot, q[i].k)-A;
	for (int i = 1; i <= m; i++) {
		if (q[i].op == 1)
			printf("%d
", Rank(q[i].l, q[i].r, q[i].k-1)+1);
		else if (q[i].op == 2) printf("%d
", kth(q[i].l, q[i].r, q[i].k));
		else if (q[i].op == 3) {
			update(q[i].l, -1);
			a[q[i].l] = q[i].k;
			update(q[i].l, 1);
		}
		else if (q[i].op == 4) {
			int g = Rank(q[i].l, q[i].r, q[i].k-1);
			if (!g) puts("-2147483647");
			else printf("%d
", kth(q[i].l, q[i].r, g));
		}
		else {
			int g = Rank(q[i].l, q[i].r, q[i].k);
			if (g == q[i].r-q[i].l+1) puts("2147483647");
			else printf("%d
", kth(q[i].l, q[i].r, g+1)); 
		}
	}
	return 0;
}