主要学习的blog/
整理下可持久化并查集要完成的内容：
1、将一个点的父节点进行更改 （实际上是新增一个信息点）update。
2、查找某个时间点下一个pos对应的信息点编号。
3、查找一个点的父节点。
4、更新一个点的deep值，不用新写函数，可以用3号操作找到编号后++。
5、初始化build。
// #pragma GCC optimize(2)
// #pragma GCC optimize(3)
// #pragma GCC optimize(4)
#include <algorithm>
#include <iterator>
#include <iostream>
#include <cstring>
#include <cstdlib>
#include <iomanip>
#include <bitset>
#include <cctype>
#include <cstdio>
#include <string>
#include <vector>
#include <stack>
#include <cmath>
#include <queue>
#include <list>
#include <map>
#include <set>
#include <cassert>
#include <unordered_map>
// #include<bits/extc++.h>
// using namespace __gnu_pbds;
using namespace std;
#define pb push_back
#define fi first
#define se second
#define debug(x) cerr<<#x << " := " << x << endl;
#define bug cerr<<"-----------------------"<<endl;
#define FOR(a, b, c) for(int a = b; a <= c; ++ a)
typedef long long ll;
typedef unsigned long long ull;
typedef long double ld;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
const int inf = 0x3f3f3f3f;
const ll inff = 0x3f3f3f3f3f3f3f3f;
const int mod = 998244353;
template<typename T>
inline T read(T&x){
x=0;int f=0;char ch=getchar();
while (ch<'0'||ch>'9') f|=(ch=='-'),ch=getchar();
while (ch>='0'&&ch<='9') x=x*10+ch-'0',ch=getchar();
return x=f?-x:x;
}
/**********showtime************/
int n,m;
const int maxn = 2e5+9;
int tr[maxn];
struct T{
int l,r;
}tree[maxn * 30];
int fa[maxn * 30], dp[maxn * 30];
int tot = 0;
void build(int & rt, int le, int ri) {
rt = ++tot;
if(le == ri) {
fa[rt] = le;
dp[rt] = 1;
return;
}
int mid = (le + ri) >> 1;
build(tree[rt].l, le, mid);
build(tree[rt].r, mid+1, ri);
}
void update(int last, int &rt, int le, int ri, int pos, int ff) {
rt = ++tot;
if(le == ri) {
fa[rt] = ff;
dp[rt] = dp[last];
return;
}
int mid = (le + ri) >> 1;
tree[rt] = tree[last];
if(mid >= pos) update(tree[last].l, tree[rt].l, le, mid, pos, ff);
else update(tree[last].r, tree[rt].r, mid+1, ri, pos, ff);
}
int query(int rt, int le, int ri, int pos) {
if(le == ri) return rt;
int mid = (le + ri) >> 1;
if(mid >= pos) return query(tree[rt].l, le, mid, pos);
else return query(tree[rt].r, mid+1, ri, pos);
}
int findfa(int rt, int pos) {
int x = query(rt, 1, n, pos);
if(fa[x] == pos) return x;
return findfa(rt, fa[x]);
}
// void adddeep()
int main(){
scanf("%d%d", &n, &m);
build(tr[0], 1, n);
for(int i=1; i<=m; i++) {
int op; scanf("%d", &op);
tr[i] = tr[i-1];
if(op == 1) {
int u,v;
scanf("%d%d", &u, &v);
int fu = findfa(tr[i], u);
int fv = findfa(tr[i], v);
if(fa[fu] == fa[fv]) {continue;}
if(dp[fu] > dp[fv]) swap(fu, fv);
update(tr[i-1], tr[i], 1, n, fa[fu], fa[fv]);
if(dp[fu] == dp[fv]) {
dp[fv] ++;
}
}
else if(op == 2) {
int k; scanf("%d", &k);
tr[i] = tr[k];
}
else {
int u,v;
scanf("%d%d", &u, &v);
int fu = findfa(tr[i], u);
int fv = findfa(tr[i], v);
// cout<<fa[fu]<<" , " << fa[fv]<<endl;
if(fa[fu] == fa[fv]) puts("1");
else puts("0");
}
}
return 0;
}
下面再附上一个可撤销并查集的模板
/// 可撤回并查集模板
struct UFS {
stack<pair<int*, int>> stk;
int fa[N], rnk[N];
inline void init(int n) {
for (int i = 0; i <= n; ++i) fa[i] = i, rnk[i] = 0;
}
inline int Find(int x) {
while(x^fa[x]) x = fa[x];
return x;
}
inline void Merge(int x, int y) {
x = Find(x), y = Find(y);
if(x == y) return ;
if(rnk[x] <= rnk[y]) {
stk.push({fa+x, fa[x]});
fa[x] = y;
if(rnk[x] == rnk[y]) {
stk.push({rnk+y, rnk[y]});
rnk[y]++;
}
}
else {
stk.push({fa+y, fa[y]});
fa[y] = x;
}
}
inline void Undo() {
*stk.top().fi = stk.top().se;
stk.pop();
}
}T;