poj 3481double queue [treap 트 리 입문 문제]

Description
The new founded Balkan Investment Group Bank (BIG-Bank) opened a new office in Bucharest, equipped with a modern computing environment provided by IBM Romania, and using modern information technologies. As usual, each client of the bank is identified by a positive integer K and, upon arriving to the bank for some services, he or she receives a positive integer priority P. One of the inventions of the young managers of the bank shocked the software engineer of the serving system. They proposed to break the tradition by sometimes calling the serving desk with the lowest priority instead of that with the highest priority. Thus, the system will receive the following types of request:
0
The system needs to stop serving
1 K P
Add client K to the waiting list with priority P
2
Serve the client with the highest priority and drop him or her from the waiting list
3
Serve the client with the lowest priority and drop him or her from the waiting list
Your task is to help the software engineer of the bank by writing a program to implement the requested serving policy.
Input
Each line of the input contains one of the possible requests; only the last line contains the stop-request (code 0). You may assume that when there is a request to include a new client in the list (code 1), there is no other request in the list of the same client or with the same priority. An identifier K is always less than 106, and a priority P is less than 107. The client may arrive for being served multiple times, and each time may obtain a different priority.
Output
For each request with code 2 or 3, the program has to print, in a separate line of the standard output, the identifier of the served client. If the request arrives when the waiting list is empty, then the program prints zero (0) to the output.
Sample Input

2
1 20 14
1 30 3
2
1 10 99
3
2
2
0

Sample Output

0
20
30
10
0

Source
Southeastern Europe 2007
9 월 에 공 대 대신 이 와 서 강 의 를 했 는데 최면 에 걸 린 것 처럼 반 시간 동안 혼수 상태 에 빠 졌 고 지연 증 말기 환 자 는 마침내 균형 수 를 다시 공부 하기 로 결심 했다.몰라 보 니 깜짝 놀 랐 다.원래 treap 트 리, splay 트 리 는 모두 밸 런 스 트 리 입 니 다. 그것 도 3 대 밸 런 스 트 리 가 있 고, 다른 하 나 는 SBT 다.
이해 하기 쉬 운 나무 짓 기!플러그 dp 에 비해 의장 트 리 라 는 말 이 있 습 니 다.그리고 treap 트 리 도 좌우 아래 에 저 장 된 다음 그룹의 아래 번호, 번호 세트 번호, 번호 세트 번호 ~ ~ ~

/***************
poj3481
2016.1.26
2052K	235MS	C++	2518B
***************/
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <cstdlib>
#include <cmath>
#include <utility>
#include <ctime>
#define max(x,y) ((x)>(y)?(x):(y))
#define min(x,y) ((x)>(y)?(y):(x))
#define INF 0x3f3f3f3f
#define MAXN 100005

using namespace std;

int cnt=1,rt=0; //     1  

struct Tree
{
  int key, size, pri, son[2],num; //     pri     pri
  void set(int x, int y, int z,int a)
  {
    key=x;
    pri=y;
    size=z;
    num=a;
    son[0]=son[1]=0;
  }
}T[MAXN];

void rotate(int p, int &x)
{
  int y=T[x].son[!p];
  T[x].size=T[x].size-T[y].size+T[T[y].son[p]].size;
  T[x].son[!p]=T[y].son[p];
  T[y].size=T[y].size-T[T[y].son[p]].size+T[x].size;
  T[y].son[p]=x;
  x=y;
}

void ins(int key, int &x,int a)
{
  if(x == 0)
    T[x = cnt++].set(key, rand(), 1,a);
  else
  {
    T[x].size++;
    int p=key < T[x].key;
    ins(key, T[x].son[!p],a);///    ，    
    if(T[x].pri < T[T[x].son[!p]].pri)///                
      rotate(p, x);
  }
}

void del(int key, int &x) //    key   
{
  if(T[x].key == key)
  {
    if(T[x].son[0] && T[x].son[1])
    {
      int p=T[T[x].son[0]].pri > T[T[x].son[1]].pri;
      rotate(p, x);
      del(key, T[x].son[p]);
    }
    else
    {
      if(!T[x].son[0])
        x=T[x].son[1];
      else
        x=T[x].son[0];
    }
  }
  else
  {
    T[x].size--;
    int p=T[x].key > key;
    del(key, T[x].son[!p]);
  }
}

int find(int p, int x) //   p            
{
  if(p == T[T[x].son[0]].size+1)
    return x;
  if(p > T[T[x].son[0]].size+1)
    find(p-T[T[x].son[0]].size-1, T[x].son[1]);
  else
    find(p, T[x].son[0]);
}

int main()
{
    //freopen("cin.txt","r",stdin);
    int que,k,p,tot=0;
    while(~scanf("%d",&que))
    {
        if(que==0) break;
        if(que==1) {
            scanf("%d%d",&k,&p);
            ins(p,rt,k);
            tot++;
        }
        if(que==2)//del max
        {
            if(tot==0)
            {
                printf("0
");
                continue;
            }
            int pos=find(tot,rt);

            printf("%d
",T[pos].num);
           // pos=1;
            del(T[pos].key,rt);
            tot--;
        }
        if(que==3)//del min
        {
            if(tot==0)
            {
                printf("0
");
                continue;
            }
            int pos=find(1,rt);

            printf("%d
",T[pos].num);
           // pos=1;
            del(T[pos].key,rt);
            tot--;
        }
    }
    return 0;
}