Treap实现名次树

插入，删除，查找的期望时间复杂度O(logn)；

//基于Treap实现的名次树，可以查第k大的数、某个数的rank
struct Node {
    Node *ch[2];
    int r, v, s;
    Node(int val = 0)
    {
        ch[0] = ch[1] = NULL;
        r = rand();
        v = val;
        s = 1;
    }
    int cmp(int x)const {
        if (x == v)return -1;
        return x < v ? 0 : 1;
    }
    void maintain()
    {
        s = 1;
        if (ch[0] != NULL)s += ch[0]->s;
        if (ch[1] != NULL)s += ch[1]->s;
    }
};
void rotate(Node* &o, int d) {
    Node * k = o->ch[1 ^ d]; //未来的顶点
    o->ch[1 ^ d] = k->ch[d]; //k的孩子变为顶点的孩子
    k->ch[d] = o;
    o->maintain();
    k->maintain();
    o = k;   //k变为顶点
}
void insert(Node* &o, int x) {
    if (o == NULL) { o = new Node(x); }
    else {
        int d = x < o->v ? 0 : 1;//相同值当作右节点插入
        insert(o->ch[d], x); if (o->ch[d]->r > o->r)rotate(o, d ^ 1);
    }
    o->maintain();
}
void remove(Node* &o, int x)
{
    int d = o->cmp(x);
    if (d == -1) {
        if (o->ch[0] == NULL)o = o->ch[1];
        else if (o->ch[1] == NULL)o = o->ch[0];
        else {
            int d2 = (o->ch[0]->r > o->ch[1]->r ? 1 : 0);
            rotate(o, d2); remove(o->ch[d2], x);
        }
    }
    else remove(o->ch[d], x);
    if (o != NULL)o->maintain();
}
int find(Node *o, int x) {
    while (o != NULL) {
        int d = o->cmp(x);
        if (d == -1)return 1;
        else o = o->ch[d];
    }
    return 0;
}
Node* root[maxn];
void init()
{
    memset(vis, 0, sizeof(vis));
    C.clear();
    for (int i = 1; i < maxn; i++)p[i] = i;
}
int find(int x)
{
    return p[x] == x ? x : p[x] = find(p[x]);
}
void mergeto(Node* &src, Node * &dest)
{
    if (src->ch[0])mergeto(src->ch[0], dest);
    if (src->ch[1])mergeto(src->ch[1], dest);
    insert(dest, src->v);
    delete src;
    src = NULL;
}
void removetree(Node* &o) 
{
    if (o->ch[0])removetree(o->ch[0]);
    if (o->ch[1])removetree(o->ch[1]);
    delete o;
    o = NULL;
}
int kth(Node* &o, int k)
{
    if (o == NULL || k <= 0 || k > o->s)return 0;
    if (o->ch[1] != NULL) {
        if (o->ch[1]->s >= k)return kth(o->ch[1], k);
        else k -= o->ch[1]->s;
    }
    if (k == 1)return o->v;
    k--;
    if (o->ch[0] != NULL) {
        return kth(o->ch[0], k); 
        if (o->ch[0]->s >= k)return kth(o->ch[0], k);
    }
}