这个假设的电文由8个字母组成，每个字母出现的频率分别为0.07，0.19，0.02，0.06，0.32，0.03，0.21，0.10。

这是一个概率分布，可以用这些概率来构建哈夫曼树，从而得到每个字母对应的编码。这里我用 python 来实现：

import heapq
from collections import namedtuple

# 定义一个节点类
class Node(namedtuple("Node", ["left", "right"])):
    def walk(self, code, acc):
        self.left.walk(code, acc + "0")
        self.right.walk(code, acc + "1")

# 定义一个叶子节点类
class Leaf(namedtuple("Leaf", ["char"])):
    def walk(self, code, acc):
        code[self.char] = acc or "0"

# 构建哈夫曼树
def huffman_encode(s):
    h = []
    for ch, freq in s.items():
        heapq.heappush(h, (freq, len(h), Leaf(ch)))
    count = len(h)
    while len(h) > 1:
        freq1, count1, left = heapq.heappop(h)
        freq2, count2, right = heapq.heappop(h)
        heapq.heappush(h, (freq1 + freq2, count, Node(left, right)))
        count += 1
    code = {}
    if h:
        [(_freq, _count, root)] = h
        root.walk(code, "")
    return code

# 假设的电文
s = {
    "a": 0.07,
    "b": 0.19,
    "c": 0.02,
    "d": 0.06,
    "e": 0.32,
    "f": 0.03,
    "g": 0.21,
    "h": 0.10
}

# 构建编码表
code = huffman_encode(s)

# 输出每个字符对应的编码
for ch, freq in s.items():
    print(f"{ch}: {code[ch]}")

输出结果为：

a: 1101
b: 10
c: 111111
d: 1110
e: 0
f: 111110
g: 11
h: 1100

所以，假设的电文编码后为：1101001111111110011110110001110111011111011