BZOJ 3224 Tyvj 1728 普通平衡树

Description

您需要写一种数据结构（可参考题目标题），来维护一些数，其中需要提供以下操作：
1. 插入x数
2. 删除x数(若有多个相同的数，因只删除一个)
3. 查询x数的排名(若有多个相同的数，因输出最小的排名)
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)

Input

第一行为n，表示操作的个数,下面n行每行有两个数opt和x，opt表示操作的序号(1<=opt<=6)

Output

对于操作3,4,5,6每行输出一个数，表示对应答案

Sample Input

10
1 106465
4 1
1 317721
1 460929
1 644985
1 84185
1 89851
6 81968
1 492737
5 493598

Sample Output

106465
84185
492737

HINT

1.n的数据范围：n<=100000
2.每个数的数据范围：[-1e7,1e7]

终于打完了splay， 不过还是全程抄代码。
%%%DQS，您是我们的红太阳！没有您我们会死！

#include<iostream>
#include<cstdio>
using namespace std;
typedef long long LL;
const int SZ = 1000010;
const int INF = 1000000010;
struct node
{
    node *ch[2], *f;
    int sz, cnt, v;

    void maintain()
    {
        sz = cnt + ch[0] -> sz + ch[1] -> sz;
    }

    int cmp(int x)
    {
        if(x == v)  return -1;
        return x < v ? 0 : 1;
    }

    int dir()
    {
        return f -> ch[1] == this;
    }

    void setc(node *x, int d)
    {
        (ch[d] = x) -> f = this;
    }

}T[SZ], *root, *null;

int Tcnt = 0;

node* newnode(int x, node *f)
{
    node *k = T + (++ Tcnt);
    k -> v = x;
    k -> ch[0] = k -> ch[1] = null;
    k -> sz = k -> cnt = 1;
    k -> f = f;
    return k;
}

void rotate(node *p)
{
    node *fa = p -> f;
    int d = p -> dir();
    fa -> f -> setc(p, fa -> dir());
    fa -> setc(p -> ch[d ^ 1], d); fa -> maintain();
    p -> setc(fa, d ^ 1); p -> maintain();
    if(fa == root) root = p;
}

void splay(node *p, node *rt = null)
{
    while(p -> f != rt)
    {
        if(p -> f -> f == rt)
            rotate(p);
        else
        {
            if(p -> dir() == p -> f -> dir())
                rotate(p -> f), rotate(p);
            else
                rotate(p), rotate(p);           
        }
    }
    p -> maintain();
}

void insert(node *p, int x)
{
    if(root == null)
    {
        root = newnode(x, null);
        return ;
    }
    while(p != null)
    {
        p -> sz ++;
        int d = p -> cmp(x);
        if(d == -1)
        {
            p -> cnt ++;
            break;
        }
        if(p -> ch[d] == null)
        {
            p -> ch[d] = newnode(x, p);
            p = p -> ch[d];
            break;
        }
        p = p -> ch[d];
    }
    splay(p);
}

void erase(node *p, int x)
{
    while(p != null)
    {
        p -> sz --;
        int d = p -> cmp(x);
        if(d == -1)
        {
            p -> cnt --; 
            break;
        }
        p = p -> ch[d];
    }
    if(p -> cnt)
        return ;
    splay(p);
    if(p -> ch[0] == null)
    {
        root = p -> ch[1];
        root -> f = null;
        return ;
    }
    if(p -> ch[1] == null)
    {
        root = p -> ch[0];
        root -> f = null;
        return ;
    }
    p = p -> ch[0];
    while(p -> ch[1] != null)   p = p -> ch[1];
    splay(p, root);
    p -> ch[1] = root -> ch[1]; p -> ch[1] -> f = p;
    p -> f = null; p -> maintain();
    root = p;
}

int ask_rank(node *p, int x)
{
    int ans = 0;
    while(p != null)
    {
        int d = p -> cmp(x);
        if(d == -1) return ans + p -> ch[0] -> sz + 1;
        if(d == 1)  ans += p -> ch[0] -> sz + p -> cnt;
        p = p -> ch[d];
    }
    return -1;
}

int ask_num(node *p, int k)
{
    while(p != null)
    {
        int l = p -> ch[0] -> sz + 1;
        int r = p -> ch[0] -> sz + p -> cnt;
        if(l <= k && k <= r)    return p -> v;
        if(k > r)   k -= r, p = p -> ch[1];
        else p = p -> ch[0];
    }
}

int ask_pre(node *p, int x)
{
    int ans = 0;
    while(p != null)
    {
        if(p -> v < x)
            ans = p -> v, p = p -> ch[1];
        else
            p = p -> ch[0];
    }
    return ans;
}

int ask_suf(node *p, int x)
{
    int ans = 0;
    while(p != null)
    {
        if(p -> v > x)
            ans = p -> v, p = p -> ch[0];
        else
            p = p -> ch[1];
    }
    return ans;
}

void init()
{
    null = newnode(-INF, null);
    null -> sz = null -> cnt = 0;
    root = null;
}

int main()
{
    init();
    int n;
    scanf("%d", &n);
    while(n--)
    {
        int op, x;
        scanf("%d%d", &op, &x);
        switch(op)
        {
            case 1 : insert(root, x); break;
            case 2 : erase(root, x); break;
            case 3 : printf("%d\n", ask_rank(root, x)); break;
            case 4 : printf("%d\n", ask_num(root, x)); break;
            case 5 : printf("%d\n", ask_pre(root, x)); break;
            case 6 : printf("%d\n", ask_suf(root, x)); break;
        }
    }
    return 0;
}