P3402 可持久化并查集

继续打打板子热热手。

#include<iostream>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<cstdlib>
#include<climits>
#include<stack>
#include<vector>
#include<queue>
#include<set>
#include<bitset>
#include<map>
//#include<regex>
#include<cstdio>
#include <iomanip>
#pragma GCC optimize(2)
#define up(i,a,b)  for(int i=a;i<b;i++)
#define dw(i,a,b)  for(int i=a;i>b;i--)
#define upd(i,a,b) for(int i=a;i<=b;i++)
#define dwd(i,a,b) for(int i=a;i>=b;i--)
//#define local
typedef long long ll;
typedef unsigned long long ull;
const double esp = 1e-6;
const double pi = acos(-1.0);
const int INF = 0x3f3f3f3f;
const int inf = 1e9;
using namespace std;
ll read()
{
	char ch = getchar(); ll x = 0, f = 1;
	while (ch<'0' || ch>'9') { if (ch == '-')f = -1; ch = getchar(); }
	while (ch >= '0' && ch <= '9') { x = x * 10 + ch - '0'; ch = getchar(); }
	return x * f;
}
typedef pair<int, int> pir;
#define lson l,mid,root<<1
#define rson mid+1,r,root<<1|1
#define lrt root<<1
#define rrt root<<1|1
const int N = 2e5 + 10;
int n, m;
struct bcj {
	int par[N * 30];
	int height[N * 30];
	int tot = 0;
	int root[N]; int ls[N * 30], rs[N * 30];
	void build(int &o,int l,int r)
	{
		o = ++tot;
		int mid = (l + r) >> 1;
		if (l == r) { par[o] = l; return; }
		build(ls[o], l, mid);
		build(rs[o], mid + 1, r);
	}
	void merge(int &o, int pre, int l, int r, int pos, int val)
	{
		o = ++tot;
		ls[o] = ls[pre];
		rs[o] = rs[pre];
		if (l == r)
		{
			height[o] = height[pre];
			par[o] = val; return;
		}
		int mid = (l + r) >> 1;
		if (pos <= mid)merge(ls[o], ls[pre], l, mid, pos, val);
		else merge(rs[o], rs[pre], mid + 1, r, pos, val);
	}
	void update(int o, int l, int r, int pos)
	{
		if (l == r)
		{
			height[o]++;
			return;
		}
		int mid = (l + r) >> 1;
		if (pos <= mid)update(ls[o], l, mid, pos);
		else update(rs[o], mid + 1, r, pos);
	}
	int query(int o, int l, int r, int pos)
	{
		if (l == r)return o;
		int mid = (l + r) >> 1;
		if (pos <= mid)return query(ls[o], l, mid, pos);
		else return query(rs[o], mid + 1, r, pos);
	}
	int find(int o,int pos)
	{
		int now = query(o, 1, n, pos);
		if (par[now] != pos)return find(o, par[now]);
		else return now;
	}
}T;
int main()
{
	n = read(), m = read();
	T.build(T.root[0], 1, n);
	int a, b, op;
	int cnt = 0;
	while (m--)
	{
		op = read();
		cnt++;
		if (op==1)
		{
			T.root[cnt] = T.root[cnt - 1];
			a = read(), b = read();
			int pos1 = T.find(T.root[cnt], a);
			int pos2 = T.find(T.root[cnt], b);
			if (T.par[pos1] == T.par[pos2])continue;
			if (T.height[pos1] < T.height[pos2])
				swap(pos1, pos2);
			T.merge(T.root[cnt], T.root[cnt - 1], 1, n, T.par[pos2], T.par[pos1]);
			if (T.height[pos1] == T.height[pos2])
			{
				T.update(T.root[cnt], 1, n, T.par[pos1]);
			}
		}
		else if (op == 2)
		{
			a = read();
			T.root[cnt] = T.root[a];
		}
		else {
			a = read(), b = read();
			T.root[cnt] = T.root[cnt - 1];
			int pos1 = T.find(T.root[cnt], a);
			int pos2 = T.find(T.root[cnt], b);
			if (T.par[pos1] == T.par[pos2])
			{
				printf("1\n");
			}
			else printf("0\n");
		}
	}
}