Poj 1442 -- black box

Black Box

Time limit:1000 ms		Memory limit:10000 K
Total submissions:7183		Accepted:2920

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 encoded 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 encoded 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 43 1 -4 2 8 -1000 21 2 6 6

Sample output

3312

Meaning: explain with test data. The number of operations is 7 and 4. The first operation is 1, indicating the first small number in the previous number, and the second operation is 2, indicating the number of 2nd small numbers in the first two,
The third operation is 6, indicating the number of the first six numbers is 3rd small. The fourth operation is 6, indicating the first 6 of the 4th small.

Idea: two heaps are used, one is the largest and the other is the smallest. The maximum heap is used for maintenance, and the minimum heap is used to calculate k decimal places.

 1 /*====================================================================== 2  *           Author :   kevin 3  *         Filename :   BlackBox.cpp 4  *       Creat time :   2014-07-28 15:46 5  *      Description : 6  ========================================================================*/ 7 #include <iostream> 8 #include <algorithm> 9 #include <cstdio>10 #include <cstring>11 #include <queue>12 #include <cmath>13 #define clr(a,b) memset(a,b,sizeof(a))14 #define M 3000515 using namespace std;16 int s[M];17 struct cmp18 {19     bool operator() (const int& x,const int& y){20         return x > y;21     }22 };23 int main(int argc,char *argv[])24 {25     int n,m;26     while(scanf("%d%d",&n,&m)!=EOF){27         priority_queue<int>max;28         priority_queue<int,vector<int>,cmp>min;29         for(int i = 0; i < n; i++){30             scanf("%d",&s[i]);31         }32         int a,t = 0,minheap,maxheap;33         for(int i = 0; i < m; i++){34             scanf("%d",&a);35             for(int i = t; i < a; i++){36                 min.push(s[i]);37                 if(!max.empty()){38                     minheap = min.top();39                     maxheap = max.top();40                     if(minheap < maxheap){41                         min.pop(); max.pop();42                         min.push(maxheap);43                         max.push(minheap);44                     }45                 }46             }47             t = a;48             printf("%d\n",min.top());49             max.push(min.top());50             min.pop();51         }52     }53     return 0;54 }

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