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连接（应该可以这么说吧）优先队列——POJ 1442

2024-07-18 21:48:04   224人阅读

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Black Box
Time Limit:1000MS     Memory Limit:10000KB     64bit IO Format:%I64d & %I64u
Submit Status

Description

Our Black Box represents a primitive database. It can save an integer array and has a special i variable. At the initial moment Black Box is empty and i equals 0. This Black Box processes a sequence of commands (transactions). There are two types of transactions: 

ADD (x): put element x into Black Box; 
GET: increase i by 1 and give an i-minimum out of all integers containing in the Black Box. Keep in mind that i-minimum is a number located at i-th place after Black Box elements sorting by non- descending. 

Let us examine a possible sequence of 11 transactions: 

Example 1 
N Transaction i Black Box contents after transaction Answer 
      (elements are arranged by non-descending)   

1 ADD(3)      0 3 
2 GET         1 3                                    3 
3 ADD(1)      1 1, 3 
4 GET         2 1, 3                                 3 
5 ADD(-4)     2 -4, 1, 3 
6 ADD(2)      2 -4, 1, 2, 3 
7 ADD(8)      2 -4, 1, 2, 3, 8 
8 ADD(-1000)  2 -1000, -4, 1, 2, 3, 8 
9 GET         3 -1000, -4, 1, 2, 3, 8                1 
10 GET        4 -1000, -4, 1, 2, 3, 8                2 
11 ADD(2)     4 -1000, -4, 1, 2, 2, 3, 8

It is required to work out an efficient algorithm which treats a given sequence of transactions. The maximum number of ADD and GET transactions: 30000 of each type. 


Let us describe the sequence of transactions by two integer arrays: 


1. A(1), A(2), ..., A(M): a sequence of elements which are being included into Black Box. A values are integers not exceeding 2 000 000 000 by their absolute value, M <= 30000. For the Example we have A=(3, 1, -4, 2, 8, -1000, 2). 

2. u(1), u(2), ..., u(N): a sequence setting a number of elements which are being included into Black Box at the moment of first, second, ... and N-transaction GET. For the Example we have u=(1, 2, 6, 6). 

The Black Box algorithm supposes that natural number sequence u(1), u(2), ..., u(N) is sorted in non-descending order, N <= M and for each p (1 <= p <= N) an inequality p <= u(p) <= M is valid. It follows from the fact that for the p-element of our u sequence we perform a GET transaction giving p-minimum number from our A(1), A(2), ..., A(u(p)) sequence. 


Input

Input contains (in given order): M, N, A(1), A(2), ..., A(M), u(1), u(2), ..., u(N). All numbers are divided by spaces and (or) carriage return characters.

Output

Write to the output Black Box answers sequence for a given sequence of transactions, one number each line.

Sample Input

7 4
3 1 -4 2 8 -1000 2
1 2 6 6

Sample Output

3
3
1
2


题意：给定M个数，每次可以插入序列一个数；再给N个数，表示在插入第几个数时输出一个数，第一次输出序列中最小的，第二次输出序列中第二小的……以此类推，直到输出N个数。


分析：因为输出时是按照先输出最小的，再输出第二小这样的方式输出的，相当于依次输出一个有序序列中的值。但因为这个序列不是固定不变的，而是不断的在更新，所以用数组是无法实现的。我们可以用优先队列来做。

定义两个优先队列，一个用来存储前k小的数，大数在前，小数在后；另一个优先队列第k+1小到最大的数，小数在前，大数在后。每次拿到一个数，先判断第一个优先队列中的数满不满k个，如果不满k个，则直接把这个数压入到第一个队列；如果满k个，判断这个数和第一个优先队列中的第一个数的大小：如果比第一个数大，就压入第二个优先队列；如果比第一个数小，就把第一个优先队列的队首元素弹出压入第二个队列，把这个新数压入第一个优先队列。

别人的题解，颇具启发性~

两个优先队列即是这样：




#include<cstdio>
#include<cstdlib>
#include<cmath>
#include<map>
#include<set>
#include<queue>
#include<stack>
#include<vector>
#include<algorithm>
#include<cstring>
#include<string>
#include<iostream>
const int MAXN=30000+10;
using namespace std;
int a[MAXN];
int b[MAXN];

int main()
{
	freopen("in.txt","r",stdin);
	int n,m;
	while(scanf("%d%d", &n,&m)==2)
	{
		priority_queue<int>q1;//栈顶元素最大
		priority_queue<int,vector<int>,greater<int> >q2;//栈顶元素最小
		int i,j;
		for(i=1; i<=n; i++){
			scanf("%d", &a[i]);
		}
		for(i=1; i<=m; i++){
			scanf("%d", &b[i]);
		}
		j=1;
		int k=1;//要输出第k小的数
		for(i=1; i<=n; i++){//把数放入栈
			if(q1.size()<k) q1.push(a[i]);//q1如果没有达到k个数，把数放入q1
			else{    //否则，用a[i]和q1队首元素比较，小的在q1，大的在q2
				int t=q1.top();
				if(a[i]>=t) q2.push(a[i]);
				else{
					q2.push(t);
					q1.pop();
					q1.push(a[i]);
				}
			}
			while(i==b[j])//在放入第b[j]个数后就要输出第k小的数，即q1队首元素
			{
				printf("%d\n",q1.top());
				k++; j++;
				if(!q2.empty()){  //q1长度+1，从q2取最小元素放入q1，继续判断是不是第b[j]次放入数字
					q1.push(q2.top());
					q2.pop();
				}
			}
		}
	}
	return 0;
}




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