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可持久化线段树

可持久化数组 (P3919)

实现 
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#include <bits/stdc++.h>
using namespace std;
const int MAXN = 1e6 + 10, MAXV = 1e8 + 10;
int n, m, rt[MAXN], a[MAXN], tr[MAXV], ls[MAXV], rs[MAXV], idx;
inline void pu(int x) { tr[x] = tr[ls[x]] + tr[rs[x]]; }
inline int clone(int x)
{
    tr[++idx] = tr[x];
    ls[idx] = ls[x];
    rs[idx] = rs[x];
    return idx;
}
void build(int &x, int l, int r)
{
    if (!x) x = ++idx;
    if (l == r)
    {
        tr[x] = a[l];
        return;
    }
    int mid = (l + r) >> 1;
    build(ls[x], l, mid);
    build(rs[x], mid + 1, r);
    pu(x);
}
int update(int x, int l, int r, int p, int v)
{
    x = clone(x);
    if (l == p && p == r)
    {
        tr[x] = v;
        return x;
    }
    int mid = (l + r) >> 1;
    if (p <= mid)
        ls[x] = update(ls[x], l, mid, p, v);
    else
        rs[x] = update(rs[x], mid + 1, r, p, v);
    pu(x);
    return x;
}
int query(int x, int l, int r, int p)
{
    if (l == p && p == r) return tr[x];
    int mid = (l + r) >> 1;
    if (p <= mid)
        return query(ls[x], l, mid, p);
    else
        return query(rs[x], mid + 1, r, p);
}
int main()
{
    ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
    cin >> n >> m;
    for (int i = 1; i <= n; i++) cin >> a[i];
    build(rt[0], 1, n);
    for (int op, p, v, num, i = 1; i <= m; i++)
    {
        cin >> v >> op >> p;
        if (op == 1)
        {
            cin >> num;
            rt[i] = update(rt[v], 1, n, p, num);
        }
        else
        {
            cout << query(rt[v], 1, n, p) << '\n';
            rt[i] = rt[v];
        }
    }
    return 0;
}