Black Box 分类： POJ 栈和队列 2015-08-05 14:07 2人阅读 评论(0) 收藏

Black Box

Time Limit: 1000MS Memory Limit: 10000K

Total Submissions: 8754 Accepted: 3599

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:

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).

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

题意:共有两种操作,一是添加一个元素,二是输出第i小的元素;

输入n,m.n代表有n个要添加的元素,m代表操作的个数,

在m个操作中a[i]=v,表示在第v次添加的时候输出第i小的元素

方法:用两个优先队列,q2从小到大出对,q1从大到小出对,q1储存的是前i-1小的数据,所以q2的队头就是第i小的数据;

#include <map>
#include <list>
#include <cmath>
#include <queue>
#include <stack>
#include <vector>
#include <string>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <iostream>
#include <algorithm>
using namespace std;
#define eps 1e-9
#define LL long long
#define PI acos(-1.0)
#define INF 0x3f3f3f3f
#define CRR fclose(stdin)
#define CWW fclose(stdout)
#define RR freopen("input.txt","r",stdin)
#pragma comment(linker, "/STACK:102400000")
#define WW freopen("output.txt","w",stdout)

const int MAX = 30010;
LL a[MAX];
int b;
int main()
{
int n,m;
while(~scanf("%d %d",&n,&m))
{
priority_queue<LL >q1;
priority_queue<LL,vector<LL>,greater<LL> >q2;

for(int i=1; i<=n; i++)
{
scanf("%lld",&a[i]);
}
int k=1;
for(int i=1; i<=m; i++)
{
scanf("%I64d",&b);
for(; k<=b; k++)
{
if(q1.empty()||q1.top()<a[k])
{
q2.push(a[k]);
}
else
{
q1.push(a[k]);
LL ans=q1.top();
q1.pop();
q2.push(ans);
}
}
LL ans=q2.top();
q2.pop();
printf("%I64d\n",ans);
q1.push(ans);
}
}
return 0;
}