BZOJ 3674 可持久化并查集加强版可持久化线段树

来源：互联网发布：手机淘宝怎么开编辑：程序博客网时间：2024/04/27 18:17

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
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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

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