当前位置 博文首页 > ttoobne:BZOJ-3674 可持久化并查集加强版 (主席树+并查集按秩合

    ttoobne:BZOJ-3674 可持久化并查集加强版 (主席树+并查集按秩合

    作者:[db:作者] 时间:2021-08-30 10:31

    题目链接
    题解:
    主席树维护并查集的 father 数组以及它的历史状态。
    我觉得比较难搞得是按秩合并,我看网上其他人的代码觉得合并的好奇怪,然后我按照自己的想法合并觉得更奇怪,但是最奇怪的是它竟然过了!我怀疑是不是数据水了,然后去 洛谷 和 acwing 交了一下,竟然都过了!
    这里由于按秩合并优化是不影响答案的正确性的,所以我合并部分的正确与否先保留。(因为我合并时会将以前的版本中的节点的秩改变)
    代码:

    #include <bits/stdc++.h>
    using namespace std;
    
    typedef long long ll;
    typedef unsigned long long ull;
    #define PI acos(-1.0)
    #define INF 0x3f3f3f3f3f3f3f3f
    #define P pair<ll, int>
    #define debug(x) cout << (#x) << ": " << (x) << " "
    #define fastio ios::sync_with_stdio(false), cin.tie(0)
    const int mod = 1e9 + 7;
    const int M = 1000000 + 10;
    const int N = 200000 + 10;
    
    int n, m;
    struct node { int ls, rs, f, dep; } tree[N<<5];
    int rt[N], cnt;
    
    void build(int &now, int l = 1, int r = n)
    {
        now = ++cnt;
        if(l == r) {
            tree[now].f = l;
            tree[now].dep = 1;
            return ;
        }
        int mid = (l + r) >> 1;
        build(tree[now].ls, l, mid);
        build(tree[now].rs, mid + 1, r);
    }
    
    int query_id(int now, int pos, int l = 1, int r = n)
    {
        if(l == r) return now;
        int mid = (l + r) >> 1;
        if(pos <= mid) return query_id(tree[now].ls, pos, l, mid);
        return query_id(tree[now].rs, pos, mid + 1, r);
    }
    
    int finds(int i, int x)
    {
        int id = query_id(rt[i], x);
        if(tree[id].f == x) return id;
        return finds(i, tree[id].f);
    }
    
    void update(int &now, int pre, int a, int b, int l = 1, int r = n)
    {
        now = ++cnt;
        if(l == r) {
            tree[now].f = b;
            tree[now].dep = tree[pre].dep;
            return ;
        }
        tree[now].ls = tree[pre].ls, tree[now].rs = tree[pre].rs;
        int mid = (l + r) >> 1;
        if(a <= mid) update(tree[now].ls, tree[pre].ls, a, b, l, mid);
        else update(tree[now].rs, tree[pre].rs, a, b, mid + 1, r);
    }
    
    void update_dep(int now, int b, int a_dep, int l = 1, int r = n)
    {
        if(l == r) {
            tree[now].dep = max(a_dep + 1, tree[now].dep);
            return ;
        }
        int mid = (l + r) >> 1;
        if(b <= mid) update_dep(tree[now].ls, b, a_dep, l, mid);
        else update_dep(tree[now].rs, b, a_dep, mid + 1, r);
    }
    
    void unions(int i, int a, int b)
    {
        int id_ra = finds(i - 1, a), id_rb = finds(i - 1, b);
        if(tree[id_ra].dep > tree[id_rb].dep) swap(id_ra, id_rb);
        update(rt[i], rt[i - 1], tree[id_ra].f, tree[id_rb].f);
        update_dep(rt[i], tree[id_rb].f, tree[id_ra].dep);
    }
    
    signed main()
    {
        scanf("%d %d", &n, &m);
        build(rt[0]);
        int op, a, b, k;
        for(int i = 1; i <= m; i ++) {
            scanf("%d", &op);
            if(op == 1) {
                scanf("%d %d", &a, &b);
                unions(i, a, b);
            } else if(op == 2) {
                scanf("%d", &k);
                rt[i] = rt[k];
            } else {
                rt[i] = rt[i - 1];
                scanf("%d %d", &a, &b);
                int id_ra = finds(i, a), id_rb = finds(i, b);
                if(tree[id_ra].f != tree[id_rb].f) printf("0\n");
                else printf("1\n");
            }
        }
        return 0;
    }
    
    /*
    
      Rejoicing in hope, patient in tribulation .
    
    */
    
    
    cs