数据结构(堆):POJ 1442 Black Box
Time Limit: 1000MS | Memory Limit: 10000K | |
Total Submissions: 10658 | Accepted: 4390 |
Description
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
Output
Sample Input
7 4
3 1 -4 2 8 -1000 2
1 2 6 6
Sample Output
3
3
1
2
水题瞬秒……
#include <iostream>
#include <cstring>
#include <cstdio>
#include <queue>
using namespace std;
const int maxn=;
int a[maxn],t[maxn],n,m;
priority_queue<int>A;
priority_queue<int,vector<int>,greater<int> >B;
int main(){
scanf("%d%d",&n,&m);
for(int i=;i<=n;i++)scanf("%d",&a[i]);
for(int i=;i<=m;i++)scanf("%d",&t[i]);
for(int i=,p=;i<=m;i++){
while(p!=t[i])B.push(a[++p]);
while(A.size()<1ul*i){
A.push(B.top());
B.pop();
}
while(B.size()&&A.top()>B.top()){
A.push(B.top());
B.push(A.top());
A.pop();B.pop();
}
printf("%d\n",A.top());
}
return ;
}
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