【洛谷】3402：【模板】可持久化并查集

P3402 【模板】可持久化并查集

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 ≤ 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
");
            else    printf("0
");
            root[i] = root[i-1];
        }
    }
    return 0;
}