假設我們有一個空序列和n個需要處理的查詢。查詢以數組queries的格式給出,格式為{query,data}。查詢可以有以下三種類型:
query = 1:將提供的資料加入到序列的末端。
query = 2:列印序列開頭的元素。然後刪除該元素。
query = 3:以升序對序列進行排序。
注意,查詢類型2和3的data總是0。
因此,若輸入是n = 9,queries = {{1, 5},{1, 4},{1, 3},{1, 2},{1, 1},{2 , 0},{3, 0},{2, 0},{3, 0}},則輸出將是5和1。
每個查詢後的序列如下所示:
要解決這個問題,我們將按照下列步驟進行:
priority_queue<int> priq 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
#讓我們來看下面的實現,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; }
9, {{1, 5}, {1, 4}, {1, 3}, {1, 2}, {1, 1}, {2, 0}, {3, 0}, {2, 0}, {3, 0}}
5 1
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