treap树---Double Queue

HDU   1908

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 KP	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 identifierK 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

题意：有两个操作数 1、2和3。1代表加入一个人，紧接着输入这个人的编号与优先级；2表示在加入的人中找到优先级最大的人，然后输出他的编号并从加入的人里剔除他；3表示在加入的人中找到优先级最小的人，然后输出他的编号并从加入的人里剔除他；

思路：使用treap树插入加入的人的编号与优先级，利用二叉树左子树小右子树大的性质，可知最左边的节点的优先级最小，最右边的节点的优先级最大；

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
using namespace std;
struct data
{
    int l,r,v,vo;
    int rnd;
}tr[1100000];
int size,root,ans2;///定义全局整型变量默认初值为0；

void rturn(int &k)
{
    int t=tr[k].l;
    tr[k].l=tr[t].r;
    tr[t].r=k;
    k=t;
}

void lturn(int &k)
{
    int t=tr[k].r;
    tr[k].r=tr[t].l;
    tr[t].l=k;
    k=t;
}

void insert(int &k,int x,int xo)
{
    if(k==0)
    {
        size++;///记录已经使用的结构体数目；
        k=size;
        tr[k].v=x;
        tr[k].vo=xo;
        tr[k].rnd=rand();
        return;
    }
    if(x>tr[k].v)
    {
        insert(tr[k].r,x,xo);
        if(tr[tr[k].r].rnd<tr[k].rnd)
            lturn(k);
    }
    else
    {
        insert(tr[k].l,x,xo);
        if(tr[tr[k].l].rnd<tr[k].rnd)
            rturn(k);
    }
}

void search_min(int &k)///最小值一定在最左的节点上；
{
    if(tr[k].l==0)
    {
        ans2=tr[k].vo;
        k=tr[k].r;
        return ;
    }
     search_min(tr[k].l);
}

void search_max(int &k)///最大值一定在最右的节点上；
{
    if(tr[k].r==0)
    {
        ans2=tr[k].vo;
        k=tr[k].l;
        return ;
    }
     search_max(tr[k].r);
}

int main()
{
    int x,k,p;
    while(scanf("%d",&x)!=EOF&&x)
    {
        ans2=0;
        if(x==1)
        {
            scanf("%d%d",&k,&p);
            insert(root,p,k);
        }
        else if(x==2)
        {
            if(root)
            search_max(root);
            printf("%d
",ans2);
        }
        else if(x==3)
        {
            if(root)
            search_min(root);
            printf("%d
",ans2);
        }
    }
    return 0;
}