treap模板

想不到我一把年纪了还要被回炉重造，感谢CP

我记得好像写过一个平衡树的了？

这次写是因为碰到作业题，是一个大号的贪心背包问题，思路不难整，但是需要特殊数据结构的加持

其实就是一个更新所有大于 x的数字要求减去一个特定的值，再在相应的位置上打标，然后搜了半天搜到了fhq treap

首先我之前写平衡树跳过了treap，所以学fhq treap不得不从treap开始看

其实也没啥难理解的，就是只要理解了这个左旋和右旋，其他其实就和平衡二叉树无异了

然后模板是借鉴一个csdn博主的，贴到这里只是为了我自己用而已

#include <bits/stdc++.h>
using namespace std;
constexpr int limit =  (3000000 + 5);//防止溢出
#define INF 0x3f3f3f3f
#define inf 0x3f3f3f3f3f
#define lowbit(i) i&(-i)//一步两步
#define EPS 1e-9
#define FASTIO  ios::sync_with_stdio(false);cin.tie(0),cout.tie(0);
#define ff(a) printf("%d\n",a );
#define pi(a, b) pair<a,b>
#define rep(i, a, b) for(ll i = a; i <= b ; ++i)
#define per(i, a, b) for(ll i = b ; i >= a  ; --i)
#define MOD 998244353
#define traverse(u) for(int i = head[u]; ~i ; i = edge[i].next)
#define FOPEN freopen("C:\\Users\\tiany\\CLionProjects\\akioi\\data.txt", "rt", stdin)
#define FOUT freopen("C:\\Users\\tiany\\CLionProjects\\akioi\\dabiao.txt", "wt", stdout)
typedef long long ll;
typedef unsigned long long ull;
char buf[1 << 23], *p1 = buf, *p2 = buf, obuf[1 << 23], *O = obuf;

inline ll read() {
#define getchar() (p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<21,stdin),p1==p2)?EOF:*p1++)
    ll sign = 1, x = 0;
    char s = getchar();
    while (s > '9' || s < '0') {
        if (s == '-')sign = -1;
        s = getchar();
    }
    while (s >= '0' && s <= '9') {
        x = (x << 3) + (x << 1) + s - '0';
        s = getchar();
    }
    return x * sign;
#undef getchar
}//快读
void print(ll x) {
    if (x / 10) print(x / 10);
    *O++ = x % 10 + '0';
}

void write(ll x, char c = 't') {
    if (x < 0)putchar('-'), x = -x;
    print(x);
    if (!isalpha(c))*O++ = c;
    fwrite(obuf, O - obuf, 1, stdout);
    O = obuf;
}
const ll mod = MOD;
ll quickPow(ll base, ll expo){
    ll ans = 1;
    while (expo){
        if(expo & 1)(ans *= base) %= mod;
        expo >>= 1;
        base = base * base;
        base %= mod;
    }
    return ans % mod;
}
ll C(ll n, ll m){
    if(n < m)return 0;
    ll x = 1, y = 1;
    if(m > n - m)m = n - m;
    rep(i ,0 , m - 1){
        x = x * (n - i) % mod;
        y = y *  (i + 1) % mod;
    }
    return x * quickPow(y, mod - 2) % mod;
}
int n, m, k;
struct node{
    int l, r;
    int key;
    int val;
    int cnt;
    int size;
}tree[limit];

void update(int rt){
    tree[rt].size = tree[tree[rt].l].size + tree[tree[rt].r].size + tree[rt].cnt;
}

int root, idx;
int new_node(int key){
    tree[++idx].key = key;
    tree[idx].val = rand();
    tree[idx].cnt = tree[idx].size = 1;
    return idx;
}

void build(){
    new_node(-INF),new_node(INF);
    root = 1,tree[1].r = 2;
    update(root);
}

//右旋
void right_rotate(int &rt){ 
    int q = tree[rt].l; 
    tree[rt].l = tree[q].r,
    tree[q].r = rt,
    rt = q;//rt再变回根
    update(tree[rt].r);
    update(rt);
}

//左旋
void left_rotate(int &rt){
    int q = tree[rt].r;
    tree[rt].r = tree[q].l;
    tree[q].l = rt;
    rt = q;
    update(tree[rt].l);
    update(rt);

}

void insert(int &rt,int key)
{
    if(!rt) rt = new_node(key);
    else if (tree[rt].key == key)tree[rt].cnt ++;
    else if (tree[rt].key > key){ 
        insert(tree[rt].l,key);
        if(tree[tree[rt].l].val > tree[rt].val) right_rotate(rt);
    }
    else{
        insert(tree[rt].r,key);
        //右大左旋
        if(tree[tree[rt].r].val > tree[rt].val) left_rotate(rt);
    }
    update(rt); 
}

void del(int &rt,int key){
    if(!rt) return ;
    if(tree[rt].key == key){ 
        if(tree[rt].cnt > 1)tree[rt].cnt --;
        else if (tree[rt].l || tree[rt].r){ 
            if(!tree[rt].r||tree[tree[rt].l].val > tree[tree[rt].r].val){
                right_rotate(rt);
                del(tree[rt].r,key);
            }
            else{
                left_rotate(rt);
                del(tree[rt].l,key);
            }
        }
        else 
            rt = 0;

    }else if (tree[rt].key > key)del(tree[rt].l,key);
    else del(tree[rt].r,key);
    update(rt);
}


int get_rank(int rt, int key){
    if (!rt) return 0;
    if (tree[rt].key == key) return tree[tree[rt].l].size + 1;
    if (tree[rt].key > key)return get_rank(tree[rt].l,key);
    return tree[tree[rt].l].size + tree[rt].cnt + get_rank(tree[rt].r,key);
}

int order_of(int rt, int rank){
    if(!rt) return INF;
    if(tree[tree[rt].l].size >= rank) return order_of(tree[rt].l,rank);

    if(tree[tree[rt].l].size + tree[rt].cnt >= rank)return tree[rt].key;
    return order_of(tree[rt].r,rank - tree[tree[rt].l].size - tree[rt].cnt);
}


int prev(int rt, int key){
    if(!rt) return -INF;
    if(tree[rt].key >= key) return ::prev(tree[rt].l,key);
    return max(tree[rt].key,::prev(tree[rt].r,key)); 

}
int next(int rt, int key){
    if(!rt) return INF;
    if(tree[rt].key <= key)return ::next(tree[rt].r,key); 
    return min(tree[rt].key,::next(tree[rt].l,key)); 
}


void solve(){
    build();
    cin>>n;
    rep(_, 1, n){
        int op, val;
        cin>>op>>val;
        if(op == 1){
            insert(root, val);
        }else if(op == 2){
            del(root, val);
        }else if(op == 3){
            cout<<(get_rank(root, val) - 1)<<endl;
        }else if(op == 4){
            cout<<(order_of(root, val + 1))<<endl;
        }else if(op == 5){
            cout<<::prev(root, val)<<endl;
        }else{
            cout<<::next(root, val)<<endl;
        }
    }
};
int32_t main() {
#ifdef LOCAL
    FOPEN;
//    FOUT;
#endif
    FASTIO
//    int kase;
//    cin>>kase;
//    while (kase--)
        solve();
    cerr << "Time elapsed: " << 1.0 * clock() / CLOCKS_PER_SEC << "s\n";
    return 0;
}