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：
- 插入xxx数
- 删除xxx数(若有多个相同的数，因只删除一个)
- 查询xxx数的排名(排名定义为比当前数小的数的个数+1+1+1。若有多个相同的数，因输出最小的排名)
- 查询排名为xxx的数
- 求xxx的前驱(前驱定义为小于xxx，且最大的数)
- 求xxx的后继(后继定义为大于xxx，且最小的数)
输入输出格式
输入格式：第一行为nnn，表示操作的个数,下面nnn行每行有两个数optoptopt和xxx，optoptopt表示操作的序号( 1≤opt≤6 1 leq opt leq 6 1≤opt≤6 )
输出格式：对于操作3,4,5,63,4,5,63,4,5,6每行输出一个数，表示对应答案
输入输出样例
输入样例#1：
复制
10 1 106465 4 1 1 317721 1 460929 1 644985 1 84185 1 89851 6 81968 1 492737 5 493598
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： n≤100000 n leq 100000 n≤100000
2.每个数的数据范围： [−107,107][-{10}^7, {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;
}