poj1442 Black Box 栈和优先队列

   题意：有n个数，按顺序加入，求加入前Gi个数时第i个最小的数是多少

思路：这里需要用到STL里的优先队列priority_queue，建一个大堆和一个小堆，若想在一个无序的序列里找第n个小的数，可以先把一个序列的n-1个数放入大堆（即假设这n-1个数是该序列里最小的），然后向小堆里push数，若小堆头元素<大堆头元素（即最小的元素比最大的元素大，说明大堆里的数还不是最小的），则交换两个数。
难点：这里如果只用一个小堆的话会超时；另外每输入的一个m数，要查询的最小的数正好是第i++个，即这里每输出一个数就把该数存入大堆里，正好使大堆里数的个数不断++（由0个不断增加），以方便下一次查询第i++个小的。





Black Box

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 7205		Accepted: 2930

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 <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <stack>
#include <queue>
#include <functional>
#include <vector>
#include <map>
//priority_queue<int> pq;
//queue<int >q;
using namespace std;

#define M 0x0f0f0f0f
#define min(a,b) (a>b?b:a)
#define max(a,b) (a>b?a:b)
int main()
{
    int m,n;
    int i,j,b,c,t;
    int a[30005];
    scanf("%d %d",&n,&m);
    priority_queue< int,vector<int>,greater<int> >small;
    priority_queue< int>large;
    for(i=0; i<n; i++)
        scanf("%d",&a[i]);
    c=0;
    for(i=0; i<m; i++)
    {
        scanf("%d",&b);
        while(c<b)
        {
            small.push(a[c]);
            if(!large.empty()&&small.top()<large.top())
            {
                t=small.top();
                small.pop();
                small.push(large.top());
                large.pop();
                large.push(t);
            }
            c++;
        }
        printf("%d\n",small.top());
        //精华！！！！
        large.push(small.top());//每次i++,让large里放一个数，使得以后每次比较时
        small.pop();           //保证small里的第一个数是第i个最小的！！！！！
    }
    return 0;
}