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
说明
1 ≤ n ≤ 105,1 ≤ m ≤ 2×105
By zky 出题人大神犇
基本和上午那道题一模一样了？？重新写一下思路清晰很多，也发现了很多细节。
一定要用启发式合并，就是维护$siz$，小的那一块合并到大的那一块去，这样保证小的每次乘2，保证了复杂度$log$级。不然会T得很惨QAQ
还有就是空间一定要开足，在并查集$find$操作里面不能路径压缩，如果压缩每次都会多$log$的空间复杂度，因为保证深度是$log$，所以不用担心时间复杂度。
合并时先找到两点的$fa$再做下一步处理，包括判断$siz$大小等。
#include<bits/stdc++.h> using namespace std; int n, m; void read(int &x) { x = 0; char ch = getchar(); while(ch > '9' || ch < '0') ch = getchar(); while(ch >= '0' && ch <= '9') { x = x * 10 + ch - '0'; ch = getchar(); } } struct Node { Node *ls, *rs; int fa, pos, siz; } pool[200005*32], *tail = pool, *root[200005], *zero; Node *newnode() { Node *nd = ++ tail; nd -> ls = zero; nd -> rs = zero; nd -> fa = nd -> pos = nd -> siz = 0; return nd; } Node *build(int l, int r) { Node *nd = newnode(); if(l == r) { nd -> fa = l; nd -> siz = 1; nd -> pos = l; return nd; } int mid = (l + r) >> 1; nd -> ls = build(l, mid); nd -> rs = build(mid + 1, r); return nd; } Node *Query(Node *nd, int l, int r, int pos) { if(l == r) return nd; int mid = (l + r) >> 1; if(pos <= mid) return Query(nd -> ls, l, mid, pos); else return Query(nd -> rs, mid + 1, r, pos); } int find(Node *nd, int x) { int a = Query(nd, 1, n, x) -> fa; if(a != x) return find(nd, a); return a; } Node *Modify(Node *nd, int l, int r, int pos, int f) { Node *nnd = newnode(); if(l == r) { nnd -> siz = nd -> siz; nnd -> ls = nd -> ls; nnd -> rs = nd -> rs; nnd -> pos = nd -> pos; nnd -> fa = f; return nnd; } int mid = (l + r) >> 1; if(pos <= mid) { nnd -> rs = nd -> rs; nnd -> ls = Modify(nd -> ls, l, mid, pos, f); } else { nnd -> ls = nd -> ls; nnd -> rs = Modify(nd -> rs, mid + 1, r, pos, f); } return nnd; } Node *Change(Node *nd, int l, int r, int pos, int siz) { Node *nnd = newnode(); if(l == r) { nnd -> fa = nd -> fa; nnd -> ls = nd -> ls; nnd -> rs = nd -> rs; nnd -> pos = nd -> pos; nnd -> siz = siz; return nnd; } int mid = (l + r) >> 1; if(pos <= mid) { nnd -> rs = nd -> rs; nnd -> ls = Change(nd -> ls, l, mid, pos, siz); } else { nnd -> ls = nd -> ls; nnd -> rs = Change(nd -> rs, mid + 1, r, pos, siz); } return nnd; } int main() { scanf("%d%d", &n, &m); zero = ++ tail; zero -> ls = zero, zero -> rs = zero, zero -> fa = 0, zero -> pos = 0, zero -> siz = 0; root[0] = build(1, n); for(int i = 1; i <= m; i ++) { int opt; read(opt); if(opt == 1) { int a, b; read(a); read(b); int u = find(root[i-1], a); int v = find(root[i-1], b); Node *uu = Query(root[i-1], 1, n, u); Node *vv = Query(root[i-1], 1, n, v); if(uu -> siz > vv -> siz) swap(uu, vv); root[i] = Modify(root[i-1], 1, n, uu -> pos, vv -> pos); root[i] = Change(root[i], 1, n, vv -> pos, vv -> siz + uu -> siz); } else if(opt == 2) { int a; read(a); root[i] = root[a]; } else { int a, b; read(a); read(b); int u = find(root[i-1], a); int v = find(root[i-1], b); if(u == v) printf("1\n"); else printf("0\n"); root[i] = root[i-1]; } } return 0; }