【模板】普通平衡树 Splay

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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求
求

输入输出格式

输入格式：

第一行为 $\leq opt \leq 6 1≤opt≤6 )$

输出格式：

对于操作

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： $\leq 100000 n≤100000$

2.每个数的数据范围： ${10}^7][−107,107]$

来源：Tyvj1728 原名：普通平衡树

在此鸣谢

Splay Tree 即可；而且好写；

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstdlib>
#include<cstring>
#include<string>
#include<cmath>
#include<map>
#include<set>
#include<vector>
#include<queue>
#include<bitset>
#include<ctime>
#include<deque>
#include<stack>
#include<functional>
#include<sstream>
//#include<cctype>
//#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<int, int> pii;
#define pi acos(-1.0)
const int N = 1005;
#define REP(i,n) for(int i=0;i<(n);i++)
typedef pair<int, int> 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;
}

【模板】普通平衡树 Splay

题目描述

输入输出格式

输入输出样例

说明

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