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H - Black Box
Time Limit:1000MS
Memory Limit:10000KB
64bit IO Format:%I64d & %I64u Submit Status
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 题意很麻烦：解释一下数据 7 4 表示给出7个数，有4个询问，下一行给出7个数，在下一行有4个询问，“1”代表从头到第一个元素中最小的值，“2”代表从头到第二个元素中第二小的值，“6”代表从头到第六个元素中第三小的值，“6”代表从头到第六个元素中第四小的值，给出的询问中 a[i] <= a[j] (i < j) ；
做法，定义两个优先队列，以大优先的p1，以小优先的p2，如果要求的是第x小的值，p1中存下(x-1)个小值，那么第x个就是p2的队首，在求第一个小的值，p1为空，求完第一个小的值后，将p2的队首放入p1，再来求第二小的值，读取给出的数（a）时，如果a大于p1的队首，那么a放入p2，否则，将a放入p1，p1的队首放入p2，保证p1的个数均比p2小，且为（x-1）个，读取完数后p2的队首就是第x小的数，输出，再把p2的队首放入p1，执行之前的操作，得到下一个要求的最小值。
用两个优先队列，分开整体的数组，得到第x小值
#include
#include
#include
#include
#include
using namespace std; #define LL __int64 priority_queue
p1 ; priority_queue
,greater
> p2 ; LL a[6000000] ; int main() { int i , j , n , m , x ; LL temp ; while(scanf("%d %d", &n, &m)!=EOF) { while( !p1.empty() ) p1.pop(); while( !p2.empty() ) p2.pop() ; for(i = 1 ; i <= n ; i++) scanf("%I64d", &a[i]); i = 1 ; while(m--) { scanf("%d", &x); for( ; i <= x ; i++) { if( p1.empty() || p1.top() < a[i] ) p2.push(a[i]); else { p1.push(a[i]); temp = p1.top() ; p1.pop() ; p2.push(temp); } } temp = p2.top(); p2.pop() ; printf("%d\n", temp); p1.push(temp); } } return 0; }
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