Black Box
Time Limit: 1000MS |
| Memory Limit: 10000K |
Total Submissions: 8710 |
| Accepted: 3576 |
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
题意：第一行输入两个正整数n,m；n表示有n个数据，m代表m次操作，第二行输入n个数据，第三行有m个整数a1,a2,a3...am，每一个整数ai都表示输入第二行数据中的前ai个数据，然后求ai个数据中第h小的数字,h的值是第三行输入的整数的顺序（例如题面上第三行的数据中 “1的h的值为1””“2的h的值为2”“6的h的值为3” “6（第二个6）的h的值为4”）
思路：定义两个优先队列，一个从大到小排列q2，一个从小到大排列q1，将前h大的数放在q1中，剩下的放在q2中，每次插入优先队列都要对q2的第一个数据进行比较，如果比q2大，就直接插入到q1中，否则将q2的第一个数据插入到q1中，然后把比较的数放在q2中，当然q2的size小于h的时候直接存在q2中，总之这样做的目的是让序列中的前h大的数在q2中。
#include<iostream>
#include<algorithm>
#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<queue>
using namespace std;
int n,m;
int a[30011],b[30011];
int cnt;
int main()
{
while(scanf("%d%d",&n,&m)!=EOF)
{
for(int i=1; i<=n; i++)
{
scanf("%d",&a[i]);
}
priority_queue<int, vector<int>, greater<int> > q1;///从小到大排列
priority_queue<int, vector<int>, less<int> > q2;///从大到小排列
while(!q1.empty())
{
q1.pop();
}
while(!q2.empty())
{
q2.pop();
}
int ans = 0;
for(int i=1; i<=m; i++)
{
scanf("%d",&cnt);
for(int j=ans+1; j<=cnt; j++)
{
if(q2.size()<i)
{
q2.push(a[j]);
}
else
{
if(a[j]<q2.top())
{
q1.push(q2.top());
q2.pop();
q2.push(a[j]);
}
else
{
q1.push(a[j]);
}
}
}
if(q2.size()<i)
{
int num = i - q2.size();
for(int k=1; k<num; k++)
{
b[k] = q1.top();
q1.pop();
}
printf("%d\n",q1.top());
for(int k=1; k<num; k++)
{
q1.push(b[k]);
}
}
else
{
printf("%d\n",q2.top());
}
ans = cnt;
if(!q1.empty())
{
q2.push(q1.top());
q1.pop();
}
}
}
return 0;
}
