Blog of RuSun

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

P3402 可持久化并查集

P3402 可持久化并查集

通过可持久化数组实现可持久化并查集。并查集不能路径压缩,要按秩合并。

查看代码
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#include <cstdio>
#include <algorithm>
using namespace std;
template <class Type>
void read (Type &x)
{
char c;
bool flag = false;
while ((c = getchar()) < '0' || c > '9')
c == '-' && (flag = true);
x = c - '0';
while ((c = getchar()) >= '0' && c <= '9')
x = (x << 1) + (x << 3) + c - '0';
flag && (x = ~x + 1);
}
template <class Type, class ...rest>
void read (Type &x, rest &...y) { read(x), read(y...); }
template <class Type>
void write (Type x)
{
x < 0 && (putchar('-'), x = ~x + 1);
x > 9 && (write(x / 10), 0);
putchar('0' + x % 10);
}
const int N = 2e5 + 10, M = 1e7 + 10;
int n, m, rt[N], idx;
struct Node { int l, r, d, p; } tr[M];
void build (int &x, int l = 1, int r = n)
{
x = ++idx;
if (l == r) return tr[x].p = l, tr[x].d = 1, void();
int mid = l + r >> 1;
build(tr[x].l, l, mid), build(tr[x].r, mid + 1, r);
}
void insert (int &x, int t, int p, int d, int l = 1, int r = n)
{
tr[++idx] = tr[x]; x = idx;
if (l == r) return tr[x].p = p, tr[x].d = d, void();
int mid = l + r >> 1;
t <= mid ? insert(tr[x].l, t, p, d, l, mid) : insert(tr[x].r, t, p, d, mid + 1, r);
}
int QueryP (int x, int t, int l = 1, int r = n)
{
if (l == r) return tr[x].p;
int mid = l + r >> 1;
return t <= mid ? QueryP(tr[x].l, t, l, mid) : QueryP(tr[x].r, t, mid + 1, r);
}
int QueryD (int x, int t, int l = 1, int r = n)
{
if (l == r) return tr[x].d;
int mid = l + r >> 1;
return t <= mid ? QueryD(tr[x].l, t, l, mid) : QueryD(tr[x].r, t, mid + 1, r);
}
int fa (int t, int x)
{
int k = QueryP(t, x);
return k == x ? x : fa(t, k);
}
int main ()
{
read(n, m);
build(rt[0]);
for (int i = 1, op; i <= m; ++i)
{
read(op);
if (op == 1)
{
int a, b; read(a, b);
rt[i] = rt[i - 1];
a = fa(rt[i], a), b = fa(rt[i], b);
if (a == b) continue;
int da = QueryD(rt[i], a), db = QueryD(rt[i], b);
if (da == db)
{
insert(rt[i], a, b, da);
insert(rt[i], b, b, db + 1);
}
else
{
if (da > db) swap(a, b), swap(da, db);
insert(rt[i], a, b, da);
}
}
else if (op == 2)
{
int k; read(k);
rt[i] = rt[k];
}
else if (op == 3)
{
int a, b; read(a, b);
rt[i] = rt[i - 1];
write(fa(rt[i], a) == fa(rt[i], b)), puts("");
}
}
return 0;
}