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板子 可持久化线段树 / 主席树

发表于 更新于 分类于 ， Valine：

将线段树或者权值线段树可持久化。

主席树

不太严谨的描述，将权值线段树可持久化。可以做区间第 K 大一类问题。

luogu P3834

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struct Seg_Tree {
    int cnt, n;
    int rt[N], sum[N<<5], ls[N<<5], rs[N<<5];
    
    inline void push_up(int suc) {
        sum[suc] = sum[ls[suc]] + sum[rs[suc]];
    }
    void build(int suc, int l, int r) {
        sum[suc] = 0;
        if (l >= r)
            return;
        int mid = (l + r) >> 1;
        build(ls[suc] = ++cnt, l, mid), build(rs[suc] = ++cnt, mid+1, r);
        push_up(suc);
    }
    void update(int suc, int pre, int l, int r, int x) {
        ls[suc] = ls[pre], rs[suc] = rs[pre], sum[suc] = sum[pre] + 1;
        if (l >= r)
            return;
        int mid = (l + r) >> 1;
        if (x <= mid)
            update(ls[suc] = ++cnt, ls[pre], l, mid, x);
        else 
            update(rs[suc] = ++cnt, rs[pre], mid+1, r, x);
    }
    int query(int u, int v, int l, int r, int k) {
        if (l >= r)
            return l;
        int t = sum[ls[v]] - sum[ls[u]], mid = (l + r) >> 1;
        if (k <= t)
            return query(ls[u], ls[v], l, mid, k);
        else 
            return query(rs[u], rs[v], mid+1, r, k - t);
    }
} tr;

inline void main() {
    read(n), read(m);
    for (int i = 1; i <= n; ++i)
        read(a[i]), b[i] = a[i];
    sort(b+1, b+n+1);
    tr.n = unique(b+1, b+n+1) - b - 1;
    tr.build(tr.rt[0] = ++tr.cnt, 1, tr.n);
    for (int i = 1; i <= n; ++i) {
        int x = lower_bound(b+1, b+tr.n+1, a[i]) - b;
        tr.update(tr.rt[i] = ++tr.cnt, tr.rt[i-1], 1, tr.n, x);
    }
    int l, r, k;
    while (m--) {
        read(l), read(r), read(k);
        printf("%d\n", b[tr.query(tr.rt[l-1], tr.rt[r], 1, tr.n, k)]);
    }
}

可持久化线段树

P3919 【模板】可持久化线段树 1（可持久化数组）

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const int N = 1e6 + 9;
const int NN = 2e7 + 9;

int n, m;
int a[N];

struct PersistableSegmentTree {
    int cnt;
    int val[NN], ls[NN], rs[NN], rt[N];
    void build(int &suc, int l, int r) {
        if (!suc) suc = ++cnt;
        if (l == r) {
            val[suc] = a[l];
            return;
        }
        int mid = (l + r) >> 1;
        build(ls[suc], l, mid), build(rs[suc], mid + 1, r);
    }
    void update(int &suc, int pre, int l, int r, int x, int k) {
        suc = ++cnt;
        ls[suc] = ls[pre], rs[suc] = rs[pre];
        if (l == r) {
            val[suc] = k;
            return;
        }
        int mid = (l + r) >> 1;
        if (x <= mid) update(ls[suc], ls[pre], l, mid, x, k);
        else update(rs[suc], rs[pre], mid + 1, r, x, k);
    }
    int query(int suc, int l, int r, int x) {
        if (l == r) return val[suc];
        int mid = (l + r) >> 1;
        if (x <= mid) return query(ls[suc], l, mid, x);
        else return query(rs[suc], mid + 1, r, x);
    }
} T;

inline void main() {
    read(n), read(m);
    for (int i = 1; i <= n; ++i) read(a[i]);
    T.build(T.rt[0], 1, n);
    int base, opt, x, y;
    for (int i = 1; i <= m; ++i) {
        read(base), read(opt);
        if (opt == 1) {
            read(x), read(y);
            T.update(T.rt[i], T.rt[base], 1, n, x, y);
        }
        else {
            read(x);
            T.rt[i] = T.rt[base];
            printf("%d\n", T.query(T.rt[i], 1, n, x));
        }
    }
}