[模板]可持久化线段树(主席树)

2019-12-11
模板

题意

求静态区间第k小

题解

离散化一下,然后先建原树,依次插入$a_i$形成n个版本

每次询问时在r和l-1两个版本里同时查询即可

调试记录

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
#include <cstdio>
#include <algorithm>
#define LS(cur) a[a[cur].ls]
#define RS(cur) a[a[cur].rs]
using namespace std;
const int maxn = 2e5 + 5;
struct T{
struct A{
int l, r, ls, rs, v;
}a[maxn * 40]; int tot;
T(){tot = 1;}
void build(int cur, int l, int r){
a[cur].l = l, a[cur].r = r;
if (l == r) return;
int mid = l + r >> 1;
build(a[cur].ls = ++tot, l, mid);
build(a[cur].rs = ++tot, mid + 1, r);
}
int upd(int t, int p){
int mid = a[t].l + a[t].r >> 1, cur;
a[cur = ++tot] = a[t];
a[cur].v++;
if (a[t].l == a[t].r) return cur;
if (p <= mid) a[cur].ls = upd(a[t].ls, p);
else a[cur].rs = upd(a[t].rs, p);
a[cur].v = LS(cur).v + RS(cur).v;
return cur;
}
int Query(int u, int v, int k){
if (a[u].l == a[u].r) return a[u].l;
if (LS(v).v - LS(u).v >= k) return Query(a[u].ls, a[v].ls, k);
else return Query(a[u].rs, a[v].rs, k - (LS(v).v - LS(u).v));
}
}t; int n, Q, a[maxn], tt[maxn], link[maxn], rt[maxn];
int main(){
scanf("%d%d", &n, &Q);
for (int i = 1; i <= n; i++) scanf("%d", a + i), tt[i] = a[i];
sort(tt + 1, tt + n + 1); t.build(1, 1, n); rt[0] = 1;
for (int i = 1; i <= n; i++){
int tmp = lower_bound(tt + 1, tt + n + 1, a[i]) - tt;
link[tmp] = a[i];
a[i] = tmp;
rt[i] = t.upd(rt[i - 1], a[i]);
}
while (Q--){
int l, r, k;
scanf("%d%d%d", &l, &r, &k);
printf("%d\n", link[t.Query(rt[l - 1], rt[r], k)]);
}
return 0;
}