【模板】普通平衡树 Splay-CFANZ编程社区

题目描述

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：

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查询排名为
求
求

输入输出格式

输入格式：

第一行为【模板】普通平衡树 Splay_数据_07

输出格式：

对于操作【模板】普通平衡树 Splay_#define_08

输入输出样例

输入样例#1：

复制

输出样例#1： 复制

106465
84185
492737

说明

时空限制：1000ms,128M

1.n的数据范围：【模板】普通平衡树 Splay_数据_09

2.每个数的数据范围：【模板】普通平衡树 Splay_#include_10

来源：Tyvj1728 原名：普通平衡树

在此鸣谢

Splay Tree 即可；而且好写；

#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
//#include
//#pragma GCC optimize(2)
using namespace std;
#define maxn 400005
#define inf 0x7fffffff
//#define INF 1e18
#define rdint(x) scanf("%d",&x)
#define rdllt(x) scanf("%lld",&x)
#define rdult(x) scanf("%lu",&x)
#define rdlf(x) scanf("%lf",&x)
#define rdstr(x) scanf("%s",x)
typedef long long  ll;
typedef unsigned long long ull;
typedef unsigned int U;
#define ms(x) memset((x),0,sizeof(x))
const long long int mod = 1e9 + 7;
#define Mod 1000000000
#define sq(x) (x)*(x)
#define eps 1e-3
typedef pair pii;
#define pi acos(-1.0)
const int N = 1005;
#define REP(i,n) for(int i=0;i<(n);i++)
typedef pair pii;
inline ll rd() {
  ll x = 0;
  char c = getchar();
  bool f = false;
  while (!isdigit(c)) {
    if (c == '-') f = true;
    c = getchar();
  }
  while (isdigit(c)) {
    x = (x << 1) + (x << 3) + (c ^ 48);
    c = getchar();
  }
  return f ? -x : x;
}

ll gcd(ll a, ll b) {
  return b == 0 ? a : gcd(b, a%b);
}
ll sqr(ll x) { return x * x; }

/*ll ans;
ll exgcd(ll a, ll b, ll &x, ll &y) {
  if (!b) {
    x = 1; y = 0; return a;
  }
  ans = exgcd(b, a%b, x, y);
  ll t = x; x = y; y = t - a / b * y;
  return ans;
}
*/

int rt, n, tot = 0;

struct node {
  int ch[2];
  int ff;
  int cnt;
  int val;
  int son;
}e[maxn<<1];

void pushup(int u) {
  e[u].son = e[e[u].ch[0]].son + e[e[u].ch[1]].son + e[u].cnt;
}

void rotate(int x) {
  int y = e[x].ff;
  int z = e[y].ff;
  int k = (e[y].ch[1] == x);
  e[z].ch[e[z].ch[1] == y] = x; e[x].ff = z;
  e[y].ch[k] = e[x].ch[k ^ 1];
  e[e[x].ch[k ^ 1]].ff = y;
  e[x].ch[k ^ 1] = y; e[y].ff = x;
  pushup(y); pushup(x);
}

void splay(int x, int aim) {
  while (e[x].ff != aim) {
    int y = e[x].ff;
    int z = e[y].ff;
    if (z != aim) {
      (e[y].ch[0] == x) ^ (e[z].ch[0] == y) ? rotate(x) : rotate(y);
    }
    rotate(x);
  }
  if (aim == 0)rt = x;
}

void ins(int x) {
  int u = rt, ff = 0;
  while (u&&e[u].val != x) {
    ff = u;
    u = e[u].ch[x > e[u].val];
  }
  if (u)e[u].cnt++;// 已经存在
  else {
    u = ++tot;// 总的节点数目
    if (ff)e[ff].ch[x > e[ff].val] = u;
    e[tot].ch[0] = 0; e[tot].ch[1] = 0;
    e[tot].ff = ff; e[tot].val = x;
    e[tot].cnt = 1; e[tot].son = 1;
  }
  splay(u, 0);
}

void Find(int x) {
  // 查找x的位置
  int u = rt;
  if (u == 0)return;
  while (e[u].ch[x > e[u].val] && x != e[u].val) {
    u = e[u].ch[x > e[u].val];
  }
  splay(u, 0);
}

int nxt(int x, int f) {
  Find(x);
  int u = rt;
  if ((e[u].val > x&&f) || (e[u].val < x && !f))return u;
  u = e[u].ch[f];
  while (e[u].ch[f ^ 1])u = e[u].ch[f ^ 1];
  return u;
}

void del(int x) {
  int last = nxt(x, 0);
  int Next = nxt(x, 1);
  splay(last, 0); splay(Next, last);
  int delt = e[Next].ch[0];
  if (e[delt].cnt > 1) {
    e[delt].cnt--; splay(delt, 0);
  }
  else e[Next].ch[0] = 0;
}

int k_th(int x) {
  // rank==x
  int u = rt;
  if (e[u].son < x)return false;
  while (1) {
    int y = e[u].ch[0];
    if (x > e[y].son + e[u].cnt) {
      x -= e[y].son + e[u].cnt;
      u = e[u].ch[1];
    }
    else if (e[y].son >= x)u = y;
    else return e[u].val;
  }
}

int main()
{
  //ios::sync_with_stdio(0);
  ins(inf); ins(-inf);
  rdint(n);
  for (int i = 0; i < n; i++) {
    int op; rdint(op);
    int x;
    if (op == 1) {
      rdint(x); ins(x);
    }
    if (op == 2) {
      rdint(x); del(x);
    }
    if (op == 3) {
      rdint(x); Find(x);
      cout << e[e[rt].ch[0]].son << endl;
    }
    if (op == 4) {
      rdint(x);
      cout << k_th(x+1) << endl;
    }
    if (op == 5) {
      rdint(x);
      cout << e[nxt(x, 0)].val << endl;
    }
    if (op == 6) {
      rdint(x);
      cout << e[nxt(x, 1)].val << endl;
    }
  }
  return 0;
}

EPFL - Fighting