Blog of RuSun

\begin {array}{c} \mathfrak {One Problem Is Difficult} \\\\ \mathfrak {Because You Don't Know} \\\\ \mathfrak {Why It Is Diffucult} \end {array}

可持久化线段树模板

用于维护历史多版本的线段树。

查看代码
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#include <cstdio>
using namespace std;
const int N = 1e6 + 10, M = 1e6 + 10;
struct Node
{
int l, r, v;
} tr[4 * N + 20 * M];
int idx;
int n, m, w[N], rt[N];
void build(int &x, int l, int r)
{
x = ++idx;
if (l == r)
{
tr[x].v = w[l];
return;
}
int mid = l + r >> 1;
build(tr[x].l, l, mid);
build(tr[x].r, mid + 1, r);
}
void modify(int &x, int l, int r, int t, int k)
{
tr[++idx] = tr[x];
x = idx;
if (l == r)
{
tr[x].v = k;
return;
}
int mid = l + r >> 1;
if (t <= mid)
modify(tr[x].l, l, mid, t, k);
else
modify(tr[x].r, mid + 1, r, t, k);
}
int query(int x, int l, int r, int t)
{
if (l == r)
return tr[x].v;
int mid = l + r >> 1;
if (t <= mid)
return query(tr[x].l, l, mid, t);
return query(tr[x].r, mid + 1, r, t);
}
int main()
{
scanf("%d%d", &n, &m);
for (int i = 1; i <= n; i++)
scanf("%d", &w[i]);
build(rt[0], 1, n);
for (int i = 1, v, op, t; i <= m; i++)
{
scanf("%d%d%d", &v, &op, &t);
rt[i] = rt[v];
if (op == 1)
{
int k;
scanf("%d", &k);
modify(rt[i], 1, n, t, k);
}
else if (op == 2)
printf("%d\n", query(rt[i], 1, n, t));
}
return 0;
}