【主席树】洛谷 P3919 【模板】可持久化线段树 1(可持久化数组)
2021-05-18 20:54:00 # ACM

题链

原来这才是可持久化的基础之基础

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
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
#include <bits/stdc++.h>
//#include <ext/rope>
//using namespace __gnu_cxx;
using namespace std;
#define LL long long
#define ll long long
#define ULL unsigned long long
#define Pair pair<LL,LL>
#define ls rt<<1
#define rs rt<<1|1
#define Pi acos(-1.0)
#define eps 1e-6
#define DBINF 1e100
#define mod 998244353
#define MAXN 1e18
#define MS 1000009

int n,m;
int a[MS];
struct node{
int l,r,val;
}p[MS<<5];
int rtpos[MS];
int tot;

int build(int l,int r){
int rt = ++tot;
if(l == r){
p[rt].val = a[l];
return rt;
}
int m = l+r>>1;
p[rt].l = build(l,m);
p[rt].r = build(m+1,r);
return rt;
}

int update(int pos,int l,int r,int lart,int val){
int rt = ++tot;
p[rt] = p[lart];
if(l == r){
p[rt].val = val;
return rt;
}
int m = l+r>>1;
if(m >= pos) p[rt].l = update(pos,l,m,p[lart].l,val);
else p[rt].r = update(pos,m+1,r,p[lart].r,val);
return rt;
}

int query(int pos,int l,int r,int rt){
if(l == r) return p[rt].val;
int m = l+r>>1;
if(m >= pos) return query(pos,l,m,p[rt].l);
else return query(pos,m+1,r,p[rt].r);
}

int main(){
ios::sync_with_stdio(false);
cin >> n >> m;
for(int i=1;i<=n;i++){
cin >> a[i];
}
rtpos[0] = build(1,n);
for(int i=1;i<=m;i++){
int vis,op,pos,val;
cin >> vis >> op >> pos;
if(op == 1){
cin >> val;
rtpos[i] = update(pos,1,n,rtpos[vis],val);
}
else if(op == 2){
rtpos[i] = rtpos[vis];
cout << query(pos,1,n,rtpos[i]) << "\n";
}
}

return 0;
}
Prev
2021-05-18 20:54:00 # ACM
Next