洛谷 P3919 可持久化数组

题目链接

题意

给数组加个时间版本，维护数组的每个版本的值

思路

用可持久化线段树维护数组即可。

代码

#include <bits/stdc++.h>
using namespace std;

#define ll long long

int root[1000005], tot = 0;
struct Node
{
    int l, r, num;
}t[(1000005<<4)+1000005*int(log2(1000005)+2)];

void build(int &rt, int l, int r)
{
    rt = ++tot;
    if(l == r)
    {
        scanf("%d",&t[rt].num);
        return;
    }
    int mid = l+r >> 1;
    if(l <= mid) build(t[rt].l, l, mid);
    if(r > mid) build(t[rt].r, mid+1, r);
}

void updata(int &x, int y, int l, int r, int pos, int num)
{
    t[++tot] = t[y];
    x = tot;
    if(l == r)
    {
        t[x].num = num;
        return;
    }
    int mid = l+r >> 1;
    if(pos <= mid) updata(t[x].l, t[y].l, l, mid, pos, num);
    if(pos > mid) updata(t[x].r, t[y].r, mid+1, r, pos, num);
}

int query(int rt, int l, int r, int pos)
{
    if(l == r) return t[rt].num;
    int mid = l+r >> 1;
    if(pos <= mid) return query(t[rt].l, l, mid, pos);
    if(pos > mid) return query(t[rt].r, mid+1, r, pos);
}

int main()
{
    int n, m;
    scanf("%d%d",&n,&m);
    build(root[0],1,n);
    for(int i = 1; i <= m; ++i)
    {
        int rt, op, id, num;
        scanf("%d%d%d",&rt,&op,&id);
        if(op == 1)
        {
            scanf("%d",&num);
            updata(root[i], root[rt], 1, n, id, num);
        }
        else
        {
            printf("%d\n",query(root[rt], 1, n, id));
            root[i] = root[rt];
        }
    }
    return 0;
}