BZOJ 3674 可持久化并查集加强版 可持久化线段树

和BZOJ 3673一样。可以翻前一篇文章。

不过内存问题省去了按大小合并。

#include <cstdio>
#include <cassert>
#include <algorithm>
using namespace std;
const int M = 200001;
class PersistantArray {
struct Node {
int v;
Node *lc, *rc;
} *a[M];
Node pool[M * 60], *C;
Node *new_node(Node *lc, Node *rc, int v) {
C->lc = lc; C->rc = rc; C->v = v; return C++;
}
int version, last, n;

Node *modify(Node *p, int l, int r, int pos, int val) {
int mid = l + r >> 1;
if (l == r)
return new_node(NULL, NULL, val);
else if (pos <= mid)
return new_node(modify(p->lc, l, mid, pos, val), p->rc, 0);
else
return new_node(p->lc, modify(p->rc, mid + 1, r, pos, val), 0);
}

int get(Node *p, int l, int r, int pos) {
int mid = l + r >> 1;
if (l == r)
return p->v;
else if (pos <= mid)
return get(p->lc, l, mid, pos);
else
return get(p->rc, mid + 1, r, pos);
}

public:
PersistantArray(int initial_capacity, int initial_data)
: C(pool), version(0), last(0), n(initial_capacity) {
a[0] = C++; a[0]->lc = a[0]->rc = a[0]; a[0]->v = initial_data;
}
void roll(int new_version) { version = new_version; }
int latest_version() { return last; }
int get(int x) { return get(a[version], 1, n, x); }
void set(int x, int y) { a[++last] = modify(a[version], 1, n, x, y); }
void set(int x, int y, bool) { modify(a[version], 1, n, x, y); }
} *fa;
int now = 0, ver[M];
int find(int x) {
int fx = fa->get(x);
if (!fx) return x;
int ans = find(fx);
fa->set(x, ans, false);
return ans;
}
void merge(int x, int y) {
x = find(x), y = find(y);
if (x == y) return;
fa->set(y, x);
now = fa->latest_version();
}

int main() {
int i, p, x, y, n, m, ans = 0;
scanf("%d%d", &n, &m);
fa = new PersistantArray(n, 0);
for (int i = 1; i <= m; i++) {
scanf("%d", &p); fa->roll(now);
switch (p) {
case 1: scanf("%d%d", &x, &y); merge(x ^ ans, y ^ ans); break;
case 2: scanf("%d", &x); now = ver[x ^ ans]; break;
case 3: scanf("%d%d", &x, &y); printf("%d\n", ans = (find(x ^ ans) == find(y ^ ans))); break;
}
ver[i] = now;
}

return 0;
}

3674: 可持久化并查集加强版

Time Limit: 15 Sec  Memory Limit: 256 MB
Submit: 1165  Solved: 415

[Submit][Status][Discuss]

Description

Description：

自从zkysb出了可持久化并查集后……

hzwer:乱写能AC，暴力踩标程

KuribohG：我不路径压缩就过了！

ndsf：暴力就可以轻松虐！

zky:……

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

请注意本题采用强制在线,所给的a,b,k均经过加密,加密方法为x = x xor lastans,lastans的初始值为0

0<n,m<=2*10^5

Input

Output

Sample Input

5 6

1 1 2

3 1 2

2 1

3 0 3

2 1

3 1 2

Sample Output

1

0

1