洛谷 P3402 【模板】可持久化并查集

任重而道远

题目描述

n个集合 m个操作

操作：

• 1 a b 合并a,b所在集合

• 2 k 回到第k次操作之后的状态(查询算作操作)

• 3 a b 询问a,b是否属于同一集合，是则输出1否则输出0

输入输出格式

输入格式：

输出格式：

输入输出样例

输入样例#1： 复制

5 6
1 1 2
3 1 2
2 0
3 1 2
2 1
3 1 2

输出样例#1： 复制

1
0
1

说明

1≤n≤1e5,1≤m≤2×1e5

By zky 出题人大神犇

AC代码：

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cstring>
using namespace std;

const int N = 2e5 + 5;
struct Node {
    int fa, dep;
    Node *ls, *rs;
}pool[N * 30], *tail = pool, *root[N];
int n, m;

Node *build (int l, int r) {
    Node *nd = ++tail;
    if (l == r) {
        nd -> fa = l;
        nd -> dep = 0;
        return nd;
    }
    int mid = l + r >> 1;
    nd -> ls = build (l, mid);
    nd -> rs = build (mid + 1, r);
    return nd;
}

Node *query (Node *rt, int o, int l, int r) {
    if (l == r) return rt;
    int mid = l + r >> 1;
    if (o <= mid) return query (rt -> ls, o, l ,mid);
    else return query (rt -> rs, o, mid + 1, r);
}

Node *find (int x, Node *nd) {
    Node *f = query (nd, x, 1, n);
    return f -> fa == x ? f : find (f -> fa, nd);
}

Node *merge (Node *rt, int x, int y, int l, int r) {
    Node *nd = ++tail;
    if (l == r) {
    	nd -> dep = rt ->dep;
        nd -> fa = y;
        return nd;
    }
    int mid = l + r >> 1;
    if (x <= mid) {
        nd -> ls = merge (rt -> ls, x, y, l, mid);
        nd -> rs = rt -> rs;
    } else {
        nd -> ls = rt -> ls;
        nd -> rs = merge (rt -> rs, x, y, mid + 1, r);
    }
    return nd;
}

void update (Node *&rt, int o, int l, int r) {
	if (l == r) {
	  rt -> dep++;
	  return ;
	}
	int mid = l + r >> 1;
	if (o <= mid) 
	  update (rt -> ls, o, l, mid);
	else 
	  update (rt -> rs, o, mid + 1, r);
	return ; 
}

int main () {
    scanf ("%d%d", &n, &m);
    root[0] = build (1, n);
    for (int i = 1; i <= m; i++) {
        int opt;
        scanf ("%d", &opt);
        if (opt == 1) {
            int a, b;
            scanf ("%d%d", &a, &b);
            Node *fa = find (a, root[i - 1]);
            Node *fb = find (b, root[i - 1]);
            if (fa -> fa == fb -> fa) {
            	root[i] = root[i - 1];
		goto Nxt;
	    }
            if (fa -> dep > fb -> dep) swap (fa, fb);
            root[i] = merge (root[i - 1], fa -> fa, fb -> fa, 1, n);
            if (fa -> dep == fb -> dep) 
              update (root[i], fb -> fa, 1, n);
        } else if (opt == 2) {
            int k;
            scanf ("%d", &k);
            root[i] = root[k];
        } else {
            int a, b;
            scanf ("%d%d", &a, &b);
            root[i] = root[i - 1];
            Node *fa = find (a, root[i]);
            Node *fb = find (b, root[i]);
            if (fa -> fa == fb -> fa) puts ("1");
            else puts ("0");
        }
        Nxt:;
    }
    return 0;
}