POJ-1442 Black Box(手写堆优化)

Black Box

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 11150		Accepted: 4554

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

 

心累到现在都没过……

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。

输出时，如果第一个优先队列里的元素个数小于k，则先把第二个优先队列里的队首元素弹出压入第一个优先队列，然后输出第一个优先队列的队首元素；如果满k个，则直接输出第一个优先队列的队首元素。

#include "bits/stdc++.h"
#define mem(a,b) memset(a,b,sizeof(a))
using namespace std;
typedef long long LL;
const int MAX=30005;
int n,m;
int a[MAX],r[MAX];
struct Que{
    int h[MAX*10];
    int n;
    void ini(){mem(h,0),n=0;}
    void heapify1(int x){
        int child=x*2,key=h[x];
        while (child<=n){
            if (child<n && h[child]<h[child+1]) child++;
            if (key<h[child]){h[x]=h[child];x=child;child=x*2;}
            else break;
        }
        h[x]=key;
    }
    void insert1(int key){
        int x=++n;
        while (x>1){
            if (key>h[x/2]) h[x]=h[x/2],x/=2;
            else break;
        }
        h[x]=key;
    }
    void del1(){
        if (n==1) n=0;
        else h[1]=h[n--],heapify1(1);
    }
    void heapify2(int x){
        int child=x*2,key=h[x];
        while (child<=n){
            if (child<n && h[child]>h[child+1]) child++;
            if (h[x]>h[child]){h[x]=h[child];x=child;child=x*2;}
            else break;
        }
        h[x]=key;
    }
    void insert2(int key){
        int x=++n;
        while (x>1){
            if (key<h[x/2]) h[x]=h[x/2],x/=2;
            else break;
        }
        h[x]=key;
    }
    void del2(){
        if (n==1) n=0;
        else h[1]=h[n--],heapify2(1);
    }
}q1,q2;
int main(){
    freopen ("box.in","r",stdin);
    freopen ("box.out","w",stdout);
    int i,j,k;
    while (~scanf("%d%d",&n,&m)){
        q1.ini();q2.ini();
        for (i=1;i<=n;i++)
         scanf("%d",a+i);
        for (j=1;j<=m;j++)
         scanf("%d",r+j);
        sort(r+1,r+m+1);
        int index(1);
        for (i=1;i<=n;i++){
            if (index>n)
             break;
            if (q1.n<index)
                q1.insert1(a[i]);
            else if (a[i]<q1.h[1]){
                int zt=q1.h[1];
                q1.del1();
                q1.insert1(a[i]);
                q2.insert2(zt);
            }
            else 
             q2.insert2(a[i]);
            while (i==r[index]){
                printf("%d
",q1.h[1]);
                if (q2.n>0){
                    q1.insert1(q2.h[1]);
                    q2.del2();
                }
                index++;
            }
        }
    }
    return 0;
}