其实就是用主席树来实现的可持久化数组：

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<string>
#include<cmath>
#include<queue>
#define ll long long
#define llu unsigned ll
using namespace std;
const int maxn=1000100;
int rt[maxn],n,m,cnt;
struct node
{
    int lc,rc,val;
}t[maxn*26];

int build(int l,int r)
{
    int p=++cnt;
    t[p].lc=t[p].rc=0;
    if(l==r)
    {
        scanf("%d",&t[p].val);
        return p;
    }
    int mid=(l+r)>>1;
    t[p].lc=build(l,mid);
    t[p].rc=build(mid+1,r);
    return p;
}

int change(int np,int pos,int val,int l,int r)
{
    int p=++cnt;
    t[p]=t[np];
    if(l==r)
    {
        t[p].val=val;
        return p;
    }
    int mid=(l+r)>>1;
    if(pos<=mid) t[p].lc=change(t[np].lc,pos,val,l,mid);
    else t[p].rc=change(t[np].rc,pos,val,mid+1,r);
    return p;
}

int ask(int p,int pos,int l,int r)
{
    if(l==r) return t[p].val;
    int mid=(l+r)>>1;
    if(pos<=mid) return ask(t[p].lc,pos,l,mid);
    else return ask(t[p].rc,pos,mid+1,r);
}

int main(void)
{
    int pos,x,y,z;
    scanf("%d%d",&n,&m);
    rt[0]=build(1,n);
    for(int i=1;i<=m;i++)
    {
        scanf("%d%d",&pos,&x);
        if(x==1)
        {
            scanf("%d%d",&y,&z);
            rt[i]=change(rt[pos],y,z,1,n);
        }
        else
        {
            scanf("%d",&y);
            printf("%d\n",ask(rt[pos],y,1,n));
            rt[i]=rt[pos];
        }
    }
    return 0;
}