query = 1：将提供的数据添加到序列的末尾。
query = 2：打印序列开头的元素。然后删除该元素。
query = 3：按升序对序列进行排序。
注意，查询类型2和3的data始终为0。
因此，如果输入是n = 9，queries = {{1, 5}，{1, 4}，{1, 3}，{1, 2}，{1, 1}，{2, 0}，{3, 0}，{2, 0}，{3, 0}}，那么输出将是5和1。
每个查询后的序列如下所示：
1：{5}2：{5, 4}3：{5, 4, 3}4：{5, 4, 3, 2}5：{5, 4, 3, 2, 1}6：{4, 3, 2, 1}，打印5。7：{1, 2, 3, 4}8：{2, 3, 4}，打印1。9：{2, 3, 4}要解决这个问题，我们将按照以下步骤进行：
priority_queue<int> priqdefine one queue qfor initialize i := 0, when i < n, update (increase i by 1), do: operation := first value of queries[i] if operation is same as 1, then: x := second value of queries[i] insert x into q otherwise when operation is same as 2, then: if priq is empty, then: print first element of q delete first element from q else: print -(top element of priq) delete top element from priq otherwise when operation is same as 3, then: while (not q is empty), do: insert (-first element of q) into priq and sort delete element from q
example让我们来看下面的实现，以便更好地理解 −
#include <bits/stdc++.h>using namespace std;void solve(int n, vector<pair<int, int>> queries){ priority_queue<int> priq; queue<int> q; for(int i = 0; i < n; i++) { int operation = queries[i].first; if(operation == 1) { int x; x = queries[i].second; q.push(x); } else if(operation == 2) { if(priq.empty()) { cout << q.front() << endl; q.pop(); } else { cout << -priq.top() << endl; priq.pop(); } } else if(operation == 3) { while(!q.empty()) { priq.push(-q.front()); q.pop(); } } }}int main() { int n = 9; vector<pair<int, int>> queries = {{1, 5}, {1, 4}, {1, 3}, {1, 2}, {1, 1}, {2, 0}, {3, 0}, {2, 0}, {3, 0}}; solve(n, queries); return 0;}
输入9, {{1, 5}, {1, 4}, {1, 3}, {1, 2}, {1, 1}, {2, 0}, {3, 0}, {2, 0}, {3, 0}}
输出51
以上就是使用c++对序列执行特定操作的详细内容。