可持久化线段树 2【模板】【主席树】

>Link

>Description

给定 $n$ 个整数构成的序列 $a$，将对于指定的闭区间 $[l,r]$ 查询其区间内的第 $k$ 小值。

>解题思路

同样是一道主席树模板题。同样发现是没打博客

>代码

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#define N 200010
using namespace std;

struct node
{
	int ls, rs, sum;
} t[3600010];
int cnt, n, m, a[N], b[N], T, root[N];

void build (int l, int r)
{
	t[++cnt] = (node){0, 0, 0};
	if (l == r) return;
	int mid = (l + r) / 2;
	build (l, mid);
	build (mid + 1, r);
}
void insert (int pre, int &x, int l, int r, int p)
{
	t[x = ++cnt] = t[pre];
	t[x].sum++;
	if (l == r) return;
	int mid = (l + r) / 2;
	if (p <= mid) insert (t[pre].ls, t[x].ls, l, mid, p);
	else insert (t[pre].rs, t[x].rs, mid + 1, r, p);
}
int ask (int x, int y, int l, int r, int k)
{
	if (l >= r) return r;
	int mid = (l + r) / 2, l1 = t[x].ls, l2 = t[y].ls;
	int c = t[l2].sum - t[l1].sum;
	if (c < k)
	  return ask (t[x].rs, t[y].rs, mid + 1, r, k - c);
	else return ask (t[x].ls, t[y].ls, l, mid, k);
}

int main()
{
	scanf ("%d%d", &n, &T);
	for (int i = 1; i <= n; i++)
	{
		scanf ("%d", &a[i]);
		b[i] = a[i];
	}
	sort (b + 1, b + 1 + n);
	m = unique (b + 1, b + 1 + n) - b - 1;
	build (1, m);
	for (int i = 1; i <= n; i++)
	{
		a[i] = lower_bound (b + 1, b + 1 + m, a[i]) - b;
		insert (root[i - 1], root[i], 1, m, a[i]);
	}
	int l, r, k;
	while (T--)
	{
		scanf ("%d%d%d", &l, &r, &k);
		printf ("%d\n", b[ask (root[l - 1], root[r], 1, m, k)]);
	}
	return 0;
}