POJ 3481 Double Queue（Treap模板题）

Double Queue

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 15786		Accepted: 6998

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

题目链接：POJ 3481

看评论区好像有一种叫双端堆的数据结构可以搞定这题，然而还是不会还是用Treap吧，因为Treap本身是一颗BST，因此一直往左找可以找到最小值和其id，一直往右找可以找到最大值和其id，然后按照其对应的优先值删除一下就好了。

代码：

#include <stdio.h>
#include <iostream>
#include <algorithm>
#include <cstdlib>
#include <sstream>
#include <numeric>
#include <cstring>
#include <bitset>
#include <string>
#include <deque>
#include <stack>
#include <cmath>
#include <queue>
#include <set>
#include <map>
using namespace std;
#define INF 0x3f3f3f3f
#define LC(x) (x<<1)
#define RC(x) ((x<<1)+1)
#define MID(x,y) ((x+y)>>1)
#define fin(name) freopen(name,"r",stdin)
#define fout(name) freopen(name,"w",stdout)
#define CLR(arr,val) memset(arr,val,sizeof(arr))
#define FAST_IO ios::sync_with_stdio(false);cin.tie(0);
typedef pair<int, int> pii;
typedef long long LL;
const double PI = acos(-1.0);
const int N = 1e6 + 7;
struct Treap
{
	int ls, rs, w, v, id, sz;
	int rnd;
};
Treap T[N];
int rt, tot;

void init()
{
	rt = tot = 0;
}
void pushup(int k)
{
	T[k].sz = T[T[k].ls].sz + T[T[k].rs].sz;
}
void lturn(int &k)
{
	int rs = T[k].rs;
	T[k].rs = T[rs].ls;
	T[rs].ls = k;
	T[rs].sz = T[k].sz;
	pushup(k);
	k = rs;
}
void rturn(int &k)
{
	int ls = T[k].ls;
	T[k].ls = T[ls].rs;
	T[ls].rs = k;
	T[ls].sz = T[k].sz;
	pushup(k);
	k = ls;
}
void ins(int &k, int v, int id)
{
	if (!k)
	{
		k = ++tot;
		T[k].ls = T[k].rs = 0;
		T[k].id = id;
		T[k].rnd = rand();
		T[k].v = v;
		T[k].w = 1;
		T[k].sz = 1;
	}
	else
	{
		++T[k].sz;
		if (v == T[k].v)
			++T[k].w;
		else if (v < T[k].v)
		{
			ins(T[k].ls, v, id);
			if (T[T[k].ls].rnd < T[k].rnd)
				rturn(k);
		}
		else
		{
			ins(T[k].rs, v, id);
			if (T[T[k].rs].rnd < T[k].rnd)
				lturn(k);
		}
	}
}
void del(int &k, int v)
{
	if (!k)
		return ;
	if (v == T[k].v)
	{
		if (T[k].w > 1)
		{
			--T[k].w;
			--T[k].sz;
		}
		else
		{
			if (T[k].ls * T[k].rs == 0)
				k = T[k].ls + T[k].rs;
			else if (T[T[k].ls].rnd < T[T[k].rs].rnd)
			{
				rturn(k);
				del(k, v);
			}
			else
			{
				lturn(k);
				del(k, v);
			}
		}
	}
	else if (v < T[k].v)
	{
		--T[k].sz;
		del(T[k].ls, v);
	}
	else
	{
		--T[k].sz;
		del(T[k].rs, v);
	}
}
int getMin(int k)
{
	if (!k)
		return 0;
	return T[k].ls ? getMin(T[k].ls) : k;
}
int getMax(int k)
{
	if (!k)
		return 0;
	return T[k].rs ? getMax(T[k].rs) : k;
}
int main(void)
{
	int ops, k, p;
	init();
	srand(987321654);
	while (~scanf("%d", &ops) && ops)
	{
		if (ops == 1)
		{
			scanf("%d%d", &k, &p);
			ins(rt, p, k);
		}
		else if (ops == 2)
		{
			int indx = getMax(rt);
			printf("%d\n", T[indx].id);
			del(rt, T[indx].v);
		}
		else if (ops == 3)
		{
			int indx = getMin(rt);
			printf("%d\n", T[indx].id);
			del(rt, T[indx].v);
		}
	}
	return 0;
}