poj 3481 Double Queue(SBT解法)

最新推荐文章于 2023-04-10 20:30:38 发布

Pira

最新推荐文章于 2023-04-10 20:30:38 发布

阅读量1.4k

点赞数

分类专栏： SBT平衡树文章标签： insert list output each integer less

本文链接：https://blog.youkuaiyun.com/fp_hzq/article/details/6887058

版权

SBT平衡树专栏收录该内容

4 篇文章

订阅专栏

本文介绍了一种使用双队列实现的银行服务系统，该系统可根据客户优先级高低进行服务调度，通过平衡树的数据结构实现高效的服务请求处理。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Double Queue

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 5004		Accepted: 2336

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

Sample Output

Source

Southeastern Europe 2007

题目：http://poj.org/problem?id=348 1

分析：这题可以用SBT来解，只要维护一棵平衡树，每次删掉最大或最小就行。。。

代码：

#include<cstdio>
using namespace std;
const int mm=1111111;
int L[mm],R[mm],S[mm],V[mm],K[mm];
int a,b,c,root,tt,sum;
void Right_Rotate(int &t)
{
    int k=L[t];
    L[t]=R[k];
    R[k]=t;
    S[k]=S[t];
    S[t]=S[L[t]]+S[R[t]]+1;
    t=k;
}
void Left_Rotate(int &t)
{
    int k=R[t];
    R[t]=L[k];
    L[k]=t;
    S[k]=S[t];
    S[t]=S[L[t]]+S[R[t]]+1;
    t=k;
}
void maintain(int &t,bool flag)
{
    if(flag)
        if(S[R[R[t]]]>S[L[t]])Left_Rotate(t);
        else if(S[L[R[t]]]>S[L[t]])
            Right_Rotate(R[t]),Left_Rotate(t);
        else return;
    else
        if(S[L[L[t]]]>S[R[t]])Right_Rotate(t);
        else if(S[R[L[t]]]>S[R[t]])
            Left_Rotate(L[t]),Right_Rotate(t);
        else return;
    maintain(L[t],0);
    maintain(R[t],1);
    maintain(t,0);
    maintain(t,1);
}
void Insert(int &t,int v,int k)
{
    if(t)
    {
        ++S[t];
        if(v<V[t])Insert(L[t],v,k);
        else Insert(R[t],v,k);
        maintain(t,v>=V[t]);
    }
    else
    {
        S[t=++tt]=1;
        V[t]=v;
        K[t]=k;
        L[t]=R[t]=0;
    }
}
int outmax()
{
    int p=root,t=root;
    while(R[t])p=t,t=R[t];
    if(t!=p)R[p]=L[t];
    else root=L[t];
    return K[t];
}
int outmin()
{
    int p=root,t=root;
    while(L[t])p=t,t=L[t];
    if(t!=p)L[p]=R[t];
    else root=R[t];
    return K[t];
}
int main()
{
    sum=root=tt=S[0]=0;
    while(scanf("%d",&a),a)
    {
        if(a==1)
        {
            scanf("%d%d",&b,&c);
            Insert(root,c,b);
            ++sum;
        }
        else if(sum)
        {
            if(a==2)printf("%d\n",outmax());
            else printf("%d\n",outmin());
            --sum;
        }
        else puts("0");
    }
    return 0;
}