python计算最小优先级队列代码分享

程序员文章站 2022-03-15 15:36:56

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复制代码代码如下:

# -*- coding: utf-8 -*-

class Heap(object):

@classmethod
def parent(cls, i):
"""父结点下标"""
return int((i - 1) >> 1);

@classmethod
def left(cls, i):
"""左儿子下标"""
return (i

@classmethod
def right(cls, i):
"""右儿子下标"""
return (i

class MinPriorityQueue(list, Heap):

@classmethod
def min_heapify(cls, A, i, heap_size):
"""最小堆化A[i]为根的子树"""
l, r = cls.left(i), cls.right(i)
if l least = l
else:
least = i
if r least = r
if least != i:
A[i], A[least] = A[least], A[i]
cls.min_heapify(A, least, heap_size)

def minimum(self):
"""返回最小元素，伪码如下：
HEAP-MINIMUM(A)
1 return A[1]

T(n) = O(1)
"""
return self[0]

def extract_min(self):
"""去除并返回最小元素，伪码如下：
HEAP-EXTRACT-MIN(A)
1 if heap-size[A] 2 then error "heap underflow"
3 min ← A[1]
4 A[1] ← A[heap-size[A]] // 尾元素放到第一位
5 heap-size[A] ← heap-size[A] - 1 // 减小heap-size[A]
6 MIN-HEAPIFY(A, 1) // 保持最小堆性质
7 return min

T(n) = θ(lgn)
"""
heap_size = len(self)
assert heap_size > 0, "heap underflow"
val = self[0]
tail = heap_size - 1
self[0] = self[tail]
self.min_heapify(self, 0, tail)
self.pop(tail)
return val

def decrease_key(self, i, key):
"""将i处的值减少到key，伪码如下：
HEAP-DECREASE-KEY(A, i, key)
1 if key > A[i]
2 then error "new key is larger than current key"
3 A[i] ← key
4 while i > 1 and A[PARENT(i)] > A[i] // 不是根结点且父结点更大时
5 do exchange A[i] ↔ A[PARENT(i)] // 交换两元素
6 i ← PARENT(i) // 指向父结点位置

T(n) = θ(lgn)
"""
val = self[i]
assert key self[i] = key
parent = self.parent
while i > 0 and self[parent(i)] > self[i]:
self[i], self[parent(i)] = self[parent(i)], self[i]
i = parent(i)

def insert(self, key):
"""将key插入A，伪码如下：
MIN-HEAP-INSERT(A, key)
1 heap-size[A] ← heap-size[A] + 1 // 对元素个数增加
2 A[heap-size[A]] ← +∞ // 初始新增加元素为+∞
3 HEAP-DECREASE-KEY(A, heap-size[A], key) // 将新增元素减少到key

T(n) = θ(lgn)
"""
self.append(float('inf'))
self.decrease_key(len(self) - 1, key)

if __name__ == '__main__':
import random

keys = range(10)
random.shuffle(keys)
print(keys)

queue = MinPriorityQueue() # 插入方式建最小堆
for i in keys:
queue.insert(i)
print(queue)

print('*' * 30)

for i in range(len(queue)):
val = i % 3
if val == 0:
val = queue.extract_min() # 去除并返回最小元素
elif val == 1:
val = queue.minimum() # 返回最小元素
else:
val = queue[1] - 10
queue.decrease_key(1, val) # queue[1]减少10
print(queue, val)

print([queue.extract_min() for i in range(len(queue))])

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