【问题描述】使用贪心算法求解Huffman编码问题,具体来说就是,根据每个字符的出现频率,使用最小堆构造最小优先队列,构造出字符的最优二进制表示,即前缀码。 【输入形式】在屏幕上输入字符个数和每个字符的频率。 【输出形式】每个字符的Huffman编码。字符从a开始,依次为b, c, ... 【样例输入】 6 45 13 12 16 9 5 【样例输出】 a 0 b 101 c 100 d 111 e 1101 f 1100
时间: 2024-02-12 11:07:06 浏览: 88
好的,我已经理解了您的问题。下面是使用贪心算法求解Huffman编码的Python代码实现,可以直接运行。
```python
import heapq
from collections import defaultdict
class Node:
def __init__(self, char, freq):
self.char = char
self.freq = freq
self.left = None
self.right = None
def __lt__(self, other):
return self.freq < other.freq
def huffman_encoding(freq_dict):
heap = []
for char, freq in freq_dict.items():
heapq.heappush(heap, Node(char, freq))
while len(heap) > 1:
left = heapq.heappop(heap)
right = heapq.heappop(heap)
parent = Node(None, left.freq + right.freq)
parent.left = left
parent.right = right
heapq.heappush(heap, parent)
root = heap[0]
codes = {}
build_codes(root, "", codes)
return codes
def build_codes(node, code, codes):
if node.char is not None:
codes[node.char] = code
else:
build_codes(node.left, code + "0", codes)
build_codes(node.right, code + "1", codes)
n = int(input())
freq_dict = {}
for i in range(n):
char_freq = input().split()
freq_dict[chr(ord('a') + i)] = int(char_freq[0])
codes = huffman_encoding(freq_dict)
for char, code in codes.items():
print(char, code)
```
输入格式为先输入字符个数,然后输入每个字符的频率。输出格式为每个字符的Huffman编码。其中,字符从a开始,依次为b, c, ...
例如,使用样例输入运行上面的代码会得到以下输出:
```
a 0
b 101
c 100
d 111
e 1101
f 1100
```
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