UVA - 501 Black Box (优先队列或vector)

Description

Black Box

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.

Example 

Let us examine a possible sequence of 11 transactions:

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 sequence of elements which are being included into Black Box. A values are integers not exceeding 2 000 000 000 by their absolute value, . For the Example we have A=(3, 1, -4, 2, 8, -1000, 2).

2.

: 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 is sorted in non-descending order, and for each p ( ) an inequality 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 sequence.

Input 

The first line of the input is an integer K, then a blank line followed by K datasets. There is a blank line between datasets.

Input for each dataset contains (in given order): . All numbers are divided by spaces and (or) carriage return characters.

Output 

For each dataset, write to the output Black Box answers sequence for a given sequence of transactions. The numbers can be separated with spaces and end-of-line characters. Print a blank line between datasets.

Sample Input 

1

7 4
3 1 -4 2 8 -1000 2
1 2 6 6

Sample Output 

3
3
1
2

题意：有n个数，m次询问，对于第i次询问是：在前tmp个中找第i大的数

方法一：优先队列，两个队列，一个从小到大q1，一个从大到小的q2，要求第i大的，其实就是相当于排序后截掉第i个数之后的，那么将这个截掉后的队列倒置就是个最大堆的队列，然后每次调整这个最大堆就行了，保证队首是第i大就是了，这个每次都加进去一个就可以保证是第i大的了

#include <iostream>
#include <cstring>
#include <algorithm>
#include <cstdio>
#include <queue>
using namespace std;
const int MAXN = 30010;

priority_queue<int, vector<int>, greater<int> > tmp;
priority_queue<int, vector<int>, less<int> > ans;
int n, m;
int num[MAXN];

int main() {
	int t;
	scanf("%d", &t);
	while (t--) {
		while (!tmp.empty())
			tmp.pop();
		while (!ans.empty())
			ans.pop();
		scanf("%d%d", &n, &m);
		for (int i = 0; i < n; i++)
			scanf("%d", &num[i]);
		int cur = 0;
		for (int i = 0; i < m; i++) {
			int cnt;
			scanf("%d", &cnt);
			for (; cur < cnt; cur++) 
				tmp.push(num[cur]);
			ans.push(tmp.top());
			tmp.pop();
			while (!tmp.empty() && tmp.top() < ans.top()) {
				cnt = tmp.top();
				tmp.pop();
				tmp.push(ans.top());
				ans.pop();
				ans.push(cnt);
			}
			printf("%d\n", ans.top());
		}
		if (t)
			printf("\n");
	}
	return 0;
}

方法二：vector的二分，每读一个就插入一个

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
#include <queue>
#include <vector>
using namespace std;
const int MAXN = 30010;

int n, m;
int num[MAXN];

int main() {
	int t;
	scanf("%d", &t);
	while (t--) {
		scanf("%d%d", &n, &m);
		for (int i = 0; i < n; i++)
			scanf("%d", &num[i]);
		vector<int> v;
		vector<int>::iterator it;
		v.clear();
		int tmp, cur = 0;
		int index = 0;
		for (int i = 0; i < m; i++) {
			scanf("%d", &tmp);
			while (v.size() != tmp) {
				it = lower_bound(v.begin(), v.end(), num[cur]);
				v.insert(it, num[cur++]);
			}
			printf("%d\n", v[index++]);
		}
		if (t)
			printf("\n");
	}
	return 0;
}