最小索引優先隊列(Min index priority queue)

在之前實現的最大優先隊列和最小優先隊列,他們可以分別快速訪問到隊列中最大元索和最小元素,但是他們有一 個缺點,就是沒有辦法通過索引訪問已存在于優先隊列中的對象,并更新它們。

為了實現這個目的,在優先隊列的基礎上,學習一種新的數據結構,索引優先隊列。

接下來我們以最小索引優先隊列舉列，最大優先索引隊列，有興趣可以自行實現。

實現思路

實現功能的三個重要數組

數組items，儲存無序插入的元素的數組

數組pq，假設將items儲存的元素進行從小到大排序，并且items對應的原索引依舊保持不變，而pq儲存的元素就是items排序過后元素的原索引（參照下圖可更好地理解）

數組qp，qp的元素對應pq中的索引， qp的索引對應pq中的元素（也就對應著未排序的items中的元素，與items元素的索引保持一致）
這樣刪除時只需要根據傳入的索引在qp中尋找對應的元素就可以跟快地找到了

實現的操作方法

size()獲取隊列的大小

is_empty()判斷隊列是否為空

less(x, y)對傳入的兩個索引對應當前隊列的元素進行大小比較

swap(i, j)對傳入的兩個索引對應當前隊列中的元素進行值交換

min_elem_index()獲取最小元素的索引

is_index_exist()判斷索引在當前隊列中是否對應一個元素

insert(index, item)指定索引index處插入一個元素item

delete_min_elem()刪除最小元素，并返回最小元素插入隊列時的索引

change_item(idx, itm)指定索引idx處，將該處元素替換為itm

swim()上浮排序操作，同之前堆的排序中介紹

sink()下沉排序操作，同之前堆的排序中介紹

Python代碼實現

import operatorclass IndexMinPriorityQueue:def __init__(self, length):self.items = [None for _ in range(length)]# Ascendingly sort items and memorize the item's relative index in itemsself.pq = [None] + [i if self.items[i] else None for i in range(len(self.items))]# Its index is associate with elements in pq, and also syncs with indices of list itemsself.qp = [i if self.pq[i] else None for i in range(len(self.pq))]self.N = 0def size(self):return self.Ndef is_empty(self):return self.N == 0def less(self, i, j):"""Compare the given two items in self.items"""return operator.lt(self.items[self.pq[i]], self.items[self.pq[j]])def swap(self, i, j):"""But change the position of the two items in the subsidiary pq and qp lists"""self.pq[i], self.pq[j] = self.pq[j], self.pq[i]self.qp[self.pq[i]], self.qp[self.pq[j]] = i, jdef min_elem_index(self):"""Find the minimum element's index"""return self.pq[1]def is_index_exist(self, index):"""Judge if the given index is exist in this queue"""return self.qp[index] is not Nonedef insert(self, index, item):"""Insert an element associated with the element's index in this queue"""if self.is_index_exist(index):returnself.items[index] = itemself.N += 1# Now it isn't a orderly queueself.pq[self.N] = indexself.qp[index] = self.N# swim the last element to make list pq orderedself.swim(self.N)def delete_min_elem(self):"""Delete the minimum element, and return its index"""min_index = self.pq[1]# print(f"min_ele: {self.items[min_index]}")self.swap(1, self.N)self.pq[self.N] = Noneself.qp[min_index] = Noneself.items[min_index] = Noneself.N -= 1self.sink(1)return min_indexdef change_item(self, idx, itm):"""Substitute a item which index=idx with a new item which value=itm"""self.items[idx] = itmk = self.qp[idx]self.sink(k)self.swim(k)def swim(self, index):"""Move the smaller element up; We should only change order in pq and qp"""while index > 1:# Compare the current node with its parent node, if smaller, swim upif self.less(index, int(index/2)): # Compare values in itemsself.swap(index, int(index/2)) # But swap the mapping position in pq and qpindex = int(index/2)def sink(self, index):"""Move the bigger element down; We should only change order in pq and qp"""# print(f"SINK: idx:{index} N:{self.N}")while 2*index <= self.N:index_smaller = 2*index if 2*index+1 > self.N else \(2*index if self.less(2*index, 2*index+1) else 2*index+1)# print(f"index_smaller: {index_smaller}")# print(f"index: {index}")if self.less(index, index_smaller):breakself.swap(index, index_smaller)index = index_smaller

測試代碼

if __name__ == '__main__':IMPQ = IndexMinPriorityQueue(10)IMPQ.insert(0, 'C')IMPQ.insert(1, 'E')IMPQ.insert(2, 'A')print(f"After three insert list items now is: {IMPQ.items}")# print(f"pq: {IMPQ.pq}")# print(f"qp: {IMPQ.qp}")# print(f"min_elem_index: {IMPQ.min_elem_index()}")index, item = 0, 'B'IMPQ.change_item(0, 'B')print(f"Changed the item in index[{index}] to {item}, items now is {IMPQ.items}")index, item = 0, 'V'IMPQ.change_item(0, 'V')print(f"Changed the item in index[{index}] to {item}, items now is {IMPQ.items}")while not IMPQ.is_empty():res = IMPQ.delete_min_elem()print(f"Pop the minimum element: {res}")print(f"After delete all elements , its size: {IMPQ.size()}")print(IMPQ.is_index_exist(0))

測試結果

After three insert list items now is: ['C', 'E', 'A', None, None, None, None, None, None, None] Changed the item in index[0] to B, items now is ['B', 'E', 'A', None, None, None, None, None, None, None] Changed the item in index[0] to V, items now is ['V', 'E', 'A', None, None, None, None, None, None, None] Pop the minimum element: 2 Pop the minimum element: 1 Pop the minimum element: 0 After delete all elements , its size: 0 False

這里為了測試結果直觀，直接將數組items拿出來了，注意items中的元素是不會改變位置的，實際改變的是用來定位items中元素的pq和qp數組

總結

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