
假设我们有一个空序列和n个需要处理的查询。查询以数组queries的格式给出,格式为{query,data}。查询可以有以下三种类型:
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。
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注意,查询类型2和3的data始终为0。
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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_queuepriq Define one queue q for 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 −
#includeusing namespace std; void solve(int n, vector > queries){ priority_queue priq; queue 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 > 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
5 1










