poj 1442(两个优先队列)

Black Box

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 6109		Accepted: 2476

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

Source

Northeastern Europe 1996

这个题目我刚开始设想不用两个优先队列来搞，只保存1个值之类的。后来发现把自己绕进去了。以后当思路明确且预估可以过的时候，就不要做无意义的优化吧。

left_pq这个优先队列，表示当前box中，最小的i-1个数字（i只有在调用get的时候才增加，因此get指令的时候只push，而add指令时候，如果有push则必有pop）。

right_pq这个优先队列，保存着第i小往后的所有数字。需要注意的是，这个堆需要重载比较符，它是一个小根队列。

这样的话，当遇到add指令，则需要与left_pq的最大元素比较，如果比之小，则替换之。将替换出来的数字再push回right_pq中。如果比之大，则直接push到right_pq中。

当遇到get指令时，则直接取right_pq的最小元素，然后pop之。再push到left_pq中。

如此做即可AC

提交记录：

1、Accepted！

代码：

#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#define oo 0x3f3f3f3f
#define MAXV 500 * 2
using namespace std;
int getv[30010];
int data[30010];
struct myint {
    int num;
};
bool operator < (const myint & a, const myint & b) {
    return a.num > b.num;
}
priority_queue pq_right;
priority_queue pq_left;
int main() {
    int n, m;
    scanf("%d%d", &m, &n);
    int i, j, k; 
    for (i = 0; i < m; i++) scanf("%d", data+i);
    for (i = 0; i < n; i++) scanf("%d", getv+i);
    j = 0;
    int last = -oo;
    for (i = 0; i < m; i++) {
        int left_max;
        if (!pq_left.empty() && (left_max = pq_left.top()) && data[i] < left_max) {
                pq_left.pop();
                pq_right.push((myint){left_max});
                pq_left.push(data[i]);
        }
        else {
            pq_right.push((myint){data[i]});
        }
        while (i+1 == getv[j]) {
            myint cur = pq_right.top();
            pq_right.pop();
            printf("%d\n", cur.num);
            pq_left.push(cur.num);
            j++;
        }
    }
    return 0;
}