poj1442 Black Box stl函数练习

Our Black Box represents a题意：
给定一个序列，每次插入序列中的一个数，给定另一个序列，表示在插入第几个数的时候取数，第一次取最小的，第二次取第二小的，依次，每次取数就将取出的数输出。




用一个优先队列存储前K小的数，大数在前，用另一个优先队列存储第K+1小的数到最大的数，小的在前，每次拿到一个数，判断第一个优先队列中的数满不慢K个，不满的话就插入，慢的话判断一下这个数和第一个队列中的第一个的大小关系，如果这个数打，就插入到第二个优先队列中，如果这个数小，就把第一个队列的队首插入到第二个队列中，并弹出，然后把这个新数插入。
在取的时候，如果第一个队列里的个数小于K，则先把第二个队列里的队首弹出到第一个队列中，然后取第一个队列的队首，如果够K个，则直接取队首

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

#include <iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<queue>
using namespace std;
int a[1000000];
int main()
{
    int n,m;
    scanf("%d %d",&n,&m);
    priority_queue<int,vector<int>,greater<int> >q;
    priority_queue<int,vector<int>,less<int> >d;
    for(int i=1; i<=n; i++)
        scanf("%d",&a[i]);
    int b=0;
    for(int i=1; i<=m; i++)
    {
        int c;
        scanf("%d",&c);
        for(int j=b+1; j<=c; j++)
        {
            if(q.empty()&&d.empty())
            {
                q.push(a[j]);
            }
            else
            {
                if(a[j]>=q.top())
                {
                    q.push(a[j]);
                }
                else
                {
                    d.push(a[j]);
                    q.push(d.top());
                    d.pop();
                }
            }
        }
        b=c;
        printf("%d\n",q.top());
        d.push(q.top());
        q.pop();
    }
    return 0;
}

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<queue>
using namespace std;
int main()
{
    priority_queue<int ,vector<int>,greater<int> >xiao;
    priority_queue<int>da;
    int i,c,u,n,m;
    int a[30010];
    scanf("%d %d",&n,&m);
    for(i=0; i<n; i++)
    {
        scanf("%d",&a[i]);
    }
    c=0;
    for(i=0; i<m; i++)
    {
        scanf("%d",&u);
        while(c<u)
        {
            if(da.empty()||da.top()<a[c])
                xiao.push(a[c]);
            else
            {
                da.push(a[c]);
                xiao.push(da.top());
                da.pop();
            }
            c++;
        }
        printf("%d\n",xiao.top());
        da.push(xiao.top());
        xiao.pop();
    }
    return 0;
}