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这里只有板子没有原理QWQ
可实现
1.插入 x 数
2.删除 x 数(若有多个相同的数，只删除一个)
3.查询 x 数的排名(排名定义为比当前数小的数的个数 +1)
4.查询排名为 x 的数
5.求 x 的前驱(前驱定义为小于 x，且最大的数)
6.求 x 的后继(后继定义为大于 x，且最小的数)
原题 https://www.luogu.com.cn/problem/P3369
在 Ver1.0 基础上把指针板子修正成 C++ 的类方法版本了，null 指针使用 static 静态量来处理。然后仅需要实现类的方法中包含小于号的重载就可以使用这个名次树了。另外，这里所有涉及到的名次都是 1-index 的。
#include using namespace std;template class Treap {public:Treap() {}void insert(T x) { _insert(root, x);}void erase(T x) { _erase(root, x);}int rank(T x) { return _GetRankOfVal(root, x);}T kth(int x) { return _GetValOfRank(root, x);}T pre(T x) { Node *ans = null; query_pre(root, x, ans); return ans->v;}T nxt(T x) { Node *ans = null; query_nxt(root, x, ans); return ans->v;}bool empty() { return root->sz == 0;}int size() { return root -> sz;}private:struct Node {Node *ch[2];T v;int sz, r, cnt;Node() { sz = r = cnt = 0;}Node(const T &v):v(v) { ch[0] = ch[1] = null; r=rand(); sz = cnt = 1;}bool operator < (const Node& rhs) const { return r < rhs.r;}int cmp(const T& x) const {if(!(x < v || v < x)) return -1;return v sz + ch[1] -> sz + cnt;}};static Node *null;Node *root = null;void rotate(Node* &o, const int &d) {Node *k = o->ch[d^1]; o->ch[d^1] = k->ch[d]; k->ch[d] = o;o->upd(); k->upd(); o = k;}void _insert(Node* &o, const T &x) {if (o == null) { o = new Node(x); return;}o->sz++;int d = o->cmp(x);if (d == -1) {o->cnt++; return;}_insert(o->ch[d], x);if (o->r ch[d]->r) rotate(o, d^1);o -> upd();}void _erase(Node* &o, const T &x) {if (o == null) return;int d = o->cmp(x);if (d == -1) {Node* u = o;if (o->cnt > 1) {o->cnt--; o->sz--; return;}if (o->ch[0] != null && o->ch[1] != null) {int d2 = o->ch[0]->r > o->ch[1]->r;rotate(o, d2); _erase(o->ch[d2], x);}else {if (o->ch[0] == null) o = o->ch[1]; else o = o->ch[0];delete u;}}else _erase(o->ch[d],x);if(o != null) o->upd();}int _GetRankOfVal(Node *&o, const T &x) {if (o == null) return 1;if (!(o->v < x || x v)) return o->ch[0]->sz + 1;else if (o->v ch[0]->sz + o->cnt + _GetRankOfVal(o->ch[1], x);else return _GetRankOfVal(o->ch[0], x);}T _GetValOfRank(Node *&o, const int &k) {if (o == null) return T();if (!(o->ch[0]->sz ch[0], k);else if(o->ch[0]->sz + o->cnt ch[1], k - o->ch[0]->sz - o->cnt);return o->v;}void query_pre(Node *&o, const T &x, Node *&ans) {if (o == null) return;if (o->v ch[1], x, ans);}else query_pre(o->ch[0], x, ans);}void query_nxt(Node *&o, const T &x, Node *&ans) {if (o == null) return;if (x v) { ans = o; query_nxt(o->ch[0], x, ans);}else query_nxt(o->ch[1], x, ans);}};template typename Treap::Node* Treap::null = new Node();struct AAA {int a;// AAA(int a = 0):a(a) {}bool operator < (const AAA& rhs) const {return a < rhs.a;}};int main() {#ifdef LOCALfreopen("test.in", "r", stdin);#endifint n; scanf("%d",&n); int op,y;Treap S; for(int i=0;i<n;i++) { scanf("%d%d",&op,&y);AAA x = AAA{y}; switch(op) { case 1: S.insert(x); break; case 2: S.erase(x); break; case 3: printf("%d\n",S.rank(x)); break; case 4: printf("%d\n",S.kth(y).a); break; case 5: printf("%d\n",S.pre(x).a); break; case 6: printf("%d\n",S.nxt(x).a); break; } }return 0;}