Black Box
Time limit:1000 ms |
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Memory limit:10000 K |
Total submissions:7436 |
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Accepted:3050 |
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
N, m are given, and N numbers are input in the next row, and m numbers are input in the last row. The I Number of the last row is required, the number I in the first X of the original series ,. A good detour, and then started to use sort is a variety of TLE, a lot of great gods with the Balance Tree, almost did not scare me, then ,.. Two heaps are merged, one is the maximum heap and the other is the smallest heap. Note that the two heaps are maintained. The maximum number of the first K is the minimum number, and the minimum number is the remaining number, the top of the largest heap is the minimum number of K.
# Include <iostream> # include <cstring> # include <cstdio> # include <cctype> # include <cstdlib> # include <algorithm> # include <set> # include <vector> # include <string> # include <map> # include <queue> using namespace STD; const int maxn = 30010; int A [maxn]; priority_queue <int, vector <int>, less <int> maxq; priority_queue <int, vector <int>, greater <int> minq; int main () {int n, m, X; scanf ("% d", & N, & M ); for (INT I = 0; I <n; I + +) Scanf ("% d", & A [I]); int Pos = 0; For (INT I = 1; I <= m; I ++) {scanf ("% d", & X); While (Pos <X) minq. push (A [POS ++]); // Add the new element to the smallest heap while (maxq. size () <I) // maintain the maximum heap {maxq. push (minq. top (); minq. pop () ;}while (! Minq. empty () & maxq. top ()> minq. top () // maintain two stacks, strictly guaranteeing maxq. top () <= minq. top () {int T1 = maxq. top (), T2 = minq. top (); maxq. pop (); minq. pop (); maxq. push (T2); minq. push (T1);} printf ("% d \ n", maxq. top () ;}return 0 ;}
Poj 1442-black box (priority queue)