POJ3481 SBT模板题目

Double Queue
Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 17409 Accepted: 7508
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

很裸的BST应用题，BST的改良数据结构中，尤其以SBT出众。
代码调试简单，编写简单，理解也简单，支持操作多，时间复杂度和空间复杂度也是非常优秀的。

不多说，贴代码（仅含有插入，删除，查找最值的模板）：

#include <iostream>
#include <cstdio>
#include <cmath>
#include <cstring>
#include <map>
using namespace std;

const int INF = 1e9;
const int maxn = 1e7+10;
struct SBT{
   int left,right,sz,key;
   void init(int k){
       left = right = 0;
       sz = 1;
       key = k;
   }
}tree[maxn];
int root,tot;

void left_rotate(int &x){
    int y = tree[x].right;
    tree[x].right = tree[y].left;
    tree[y].left = x;
    tree[y].sz = tree[x].sz;
    tree[x].sz = tree[tree[x].left].sz + tree[tree[x].right].sz + 1;
    x = y;
}

void right_rotate(int &x){
    int y = tree[x].left;
    tree[x].left = tree[y].right;
    tree[y].right = x;
    tree[y].sz = tree[x].sz;
    tree[x].sz = tree[tree[x].left].sz + tree[tree[x].right].sz + 1;
    x = y;
}

void maintain(int &x, bool flag){
    if (!flag){
        if (tree[tree[tree[x].left].left].sz>tree[tree[x].right].sz) right_rotate(x);
        else if (tree[tree[tree[x].left].right].sz>tree[tree[x].right].sz)
                {left_rotate(tree[x].left); right_rotate(x);}
        else return;
    }
    else {
       if (tree[tree[tree[x].right].right].sz>tree[tree[x].left].sz) left_rotate(x);
        else if (tree[tree[tree[x].right].left].sz>tree[tree[x].left].sz)
                {right_rotate(tree[x].right);left_rotate(x);}
        else return;
    }
    maintain(tree[x].left,0);
    maintain(tree[x].right,1);
    maintain(x,0);
    maintain(x,1);
}

void Insert(int &x, int key){
    if (!x){x=++tot; tree[x].init(key); return;}
    tree[x].sz++;
    if (key<tree[x].key) Insert(tree[x].left,key);
    else Insert(tree[x].right,key);
    maintain(x,key>=tree[x].key);
}

int del(int &x, int key){
   if (tree[x].key==key || (tree[x].left==0 && key<tree[x].key) || (tree[x].right==0 && key>tree[x].key)){
       int p = tree[x].key;
       if (tree[x].left == 0 || tree[x].right == 0) x = tree[x].left+tree[x].right;
       else tree[x].key = del(tree[x].left,INF);
       return p;
   }
   if (key < tree[x].key) return del(tree[x].left,key);
   else return del(tree[x].right,key);
}

int get_min(int x){
   while (tree[x].left) x = tree[x].left;
   return x;
}

int get_max(int x){
    while(tree[x].right) x=tree[x].right;
    return x;
}

int n,k,p;
map<int,int>m;
int main(){
    int f;
    root = tot = 0;
    m.clear();
    while (scanf("%d",&n) && n){
        if (n==1) {
           scanf("%d %d",&k,&p);
           m[p] = k;
           Insert(root,p);
        }
        else if (n==2) {
            f = get_max(root);
            printf("%d\n",m[tree[f].key]);
            del(root,tree[f].key);
        }
        else {
            f = get_min(root);
            printf("%d\n",m[tree[f].key]);
            del(root,tree[f].key);
        }
    }
    return 0;
}