「luogu3380」【模板】二逼平衡树（树套树）

「luogu3380」【模板】二逼平衡树（树套树）

传送门
我写的树套树——线段树套平衡树。
线段树上的每一个节点都是一棵 \(\text{FHQ Treap}\) ，然后我们就可以根据平衡树的基本操作以及线段树上区间信息可合并的性质来实现了，具体细节看代码都懂。
参考代码：

#include <algorithm>
#include <cstdlib>
#include <cstdio>
#define rg register
#define file(x) freopen(x".in", "r", stdin), freopen(x".out", "w", stdout)
using namespace std;
template < class T > inline void read(T& s) {
    s = 0; int f = 0; char c = getchar();
    while ('0' > c || c > '9') f |= c == '-', c = getchar();
    while ('0' <= c && c <= '9') s = s * 10 + c - 48, c = getchar();
    s = f ? -s : s;
}

const int _ = 50010, __ = 2000010, INF = 2147483647;

int n, m, A[_];
int tot, ch[2][__], siz[__], pri[__], val[__];

struct node {
    int rt, a, b, c;
    
    inline int Newnode(int v) { return siz[++tot] = 1, val[tot] = v, pri[tot] = rand(), tot; }
    
    inline void pushup(int p) { siz[p] = siz[ch[0][p]] + siz[ch[1][p]] + 1; }
    
    inline void split(int p, int v, int& x, int& y) {
        if (!p) { x = y = 0; return ; }
        if (val[p] <= v) return x = p, split(ch[1][p], v, ch[1][x], y), pushup(p);
        else return y = p, split(ch[0][p], v, x, ch[0][y]), pushup(p);
    }
    
    inline int merge(int x, int y) {
        if (!x || !y) return x + y;
        if (pri[x] > pri[y]) return ch[1][x] = merge(ch[1][x], y), pushup(x), x;
        else return ch[0][y] = merge(x, ch[0][y]), pushup(y), y;
    }
    
    inline void insert(int v) { split(rt, v, a, b), rt = merge(a, merge(Newnode(v), b)); }
    
    inline void erase(int v) { split(rt, v, a, c), split(a, v - 1, a, b), b = merge(ch[0][b], ch[1][b]), rt = merge(a, merge(b, c)); }
    
    inline void build(int l, int r) { for (rg int i = l; i <= r; ++i) insert(A[i]); }
    
    inline int kth(int p, int k) {
        if (siz[ch[0][p]] + 1 > k) return kth(ch[0][p], k);
        if (siz[ch[0][p]] + 1 == k) return val[p];
        if (siz[ch[0][p]] + 1 < k) return kth(ch[1][p], k - siz[ch[0][p]] - 1);
    }
    
    inline int pre(int v) { split(rt, v - 1, a, b), c = a != 0 ? kth(a, siz[a]) : -INF, rt = merge(a, b); return c; }
    
    inline int nxt(int v) { split(rt, v, a, b), c = b != 0 ? kth(b, 1) : INF, rt = merge(a, b); return c; }
    
    inline int rank(int v) { split(rt, v - 1, a, b), c = siz[a] + 1, rt = merge(a, b); return c; }
} t[_ << 2];

inline int lc(int p) { return p << 1; }

inline int rc(int p) { return p << 1 | 1; }

inline void build(int p = 1, int l = 1, int r = n) {
    t[p].build(l, r);
    if (l == r) return ;
    int mid = (l + r) >> 1;
    build(lc(p), l, mid), build(rc(p), mid + 1, r);
}

inline void update(int x, int v, int p = 1, int l = 1, int r = n) {
    t[p].erase(A[x]), t[p].insert(v);
    if (l == r) return ;
    int mid = (l + r) >> 1;
    if (x <= mid) update(x, v, lc(p), l, mid);
    else update(x, v, rc(p), mid + 1, r);
}

inline int rank(int ql, int qr, int v, int p = 1, int l = 1, int r = n) {
    if (ql <= l && r <= qr) return t[p].rank(v) - 1;
    int mid = (l + r) >> 1, res = 0;
    if (ql <= mid) res += rank(ql, qr, v, lc(p), l, mid);
    if (qr > mid) res += rank(ql, qr, v, rc(p), mid + 1, r);
    return res;
}

inline int kth(int ql, int qr, int k) {
    int l = 0, r = 100000000, res;
    while (l <= r) {
    int mid = (l + r) >> 1;
    if (rank(ql, qr, mid) + 1 <= k) res = mid, l = mid + 1; else r = mid - 1;
    }
    return res;
}

inline int pre(int ql, int qr, int v, int p = 1, int l = 1, int r = n) {
    if (ql <= l && r <= qr) return t[p].pre(v);
    int mid = (l + r) >> 1, res = -INF;
    if (ql <= mid) res = max(res, pre(ql, qr, v, lc(p), l, mid));
    if (qr > mid) res = max(res, pre(ql, qr, v, rc(p), mid + 1, r));
    return res;
}

inline int nxt(int ql, int qr, int v, int p = 1, int l = 1, int r = n) {
    if (ql <= l && r <= qr) return t[p].nxt(v);
    int mid = (l + r) >> 1, res = INF;
    if (ql <= mid) res = min(res, nxt(ql, qr, v, lc(p), l, mid));
    if (qr > mid) res = min(res, nxt(ql, qr, v, rc(p), mid + 1, r));
    return res;
}

int main() {
#ifndef ONLINE_JUDGE
    file("cpp");
#endif
    read(n), read(m);
    for (rg int i = 1; i <= n; ++i) read(A[i]);
    build();
    for (rg int opt, l, r, k, x; m--; ) {
        read(opt);
        if (opt == 1) read(l), read(r), read(k), printf("%d\n", rank(l, r, k) + 1);
        if (opt == 2) read(l), read(r), read(k), printf("%d\n", kth(l, r, k));
        if (opt == 3) read(x), read(k), update(x, k), A[x] = k;
        if (opt == 4) read(l), read(r), read(k), printf("%d\n", pre(l, r, k));
        if (opt == 5) read(l), read(r), read(k), printf("%d\n", nxt(l, r, k));
    }
    return 0;
}

标签：rt,ch,return,树套,int,res,mid,luogu3380,模板
来源： https://www.cnblogs.com/zsbzsb/p/12232249.html