以频率作为权值构造Huffman树 A,B,C,D,E,F的频率为{4,2,6,8,3,2}
时间: 2024-05-03 21:18:54 浏览: 28
首先,将频率从小到大排序,得到{2, 2, 3, 4, 6, 8}。然后,将相邻的两个节点合并,得到{4, 3, 4, 6, 8}。再次排序,得到{3, 4, 4, 6, 8}。合并相邻的两个节点,得到{7, 4, 6, 8}。再次排序,得到{4, 6, 7, 8}。最后,合并剩下的两个节点,得到根节点,其权重为25。构造的Huffman树如下所示:
```
25
/ \
13 12
/ \
6 7
/ \ / \
A B C D
/
E
\
F
```
其中,叶节点分别表示字符 A,B,C,D,E,F。从根节点到叶节点的路径上,0表示向左子树走,1表示向右子树走,即可得到每个字符对应的Huffman编码:
A: 00
B: 01
C: 10
D: 11
E: 100
F: 101
相关问题
1、已知数据六个字母及在通信中出现频率如下表: A B C D E F 0.15 0.3 0.1 0.1 0.2 0.15 把这些字母和频率作为叶子结点及权值,完成如下工作。 (1) 画出对应的Huffman树。 (2) 计算带权路径长度WPL(要有过程)。 (3) 求A、B、C、D、E、F的Huffman编码。
好的,我明白了。这是一个经典的Huffman编码问题。我将依次回答你的问题。
1. 画出对应的Huffman树。
首先,我们需要将这六个字母按照出现频率从小到大排序,得到:C、D、F、A、E、B。然后,我们将出现频率最小的两个字母(C和D)合并为一个节点,权值为它们的权值之和(0.1+0.1=0.2)。得到以下树形结构:
```
0.2
/ \
C D
```
然后,我们将出现频率次小的字母(F)与上一步合并后的节点合并为一个节点,权值为它们的权值之和(0.15+0.2=0.35)。得到以下树形结构:
```
0.35
/ \
0.2 F
/ \
C D
```
接下来,我们将出现频率第三小的字母(A)与上一步合并后的节点合并为一个节点,权值为它们的权值之和(0.35+0.15=0.5)。得到以下树形结构:
```
0.5
/ \
0.35 A
/ \
0.2 F
/ \
C D
```
然后,我们将出现频率第四小的字母(E)与上一步合并后的节点合并为一个节点,权值为它们的权值之和(0.5+0.2=0.7)。得到以下树形结构:
```
0.7
/ \
0.5 E
/ \
0.35 A
/ \
C D
```
最后,我们将剩下的字母(B)与上一步合并后的节点合并为一个节点,权值为它们的权值之和(0.7+0.3=1.0)。得到以下最终的Huffman树:
```
1.0
/ \
0.7 B
/ \
0.5 E
/ \
0.35 A
/ \
C D
```
2. 计算带权路径长度WPL(要有过程)。
带权路径长度WPL是指Huffman树中所有叶子节点的权值乘以它们到根节点的路径长度之和。我们可以用以下公式来计算WPL:
WPL = ∑(叶子节点权值 × 叶子节点到根节点的路径长度)
根据上面的公式,我们可以列出以下表格:
| 字母 | 频率 | 编码 | 路径长度 | 权值 × 路径长度 |
| --- | --- | --- | --- | --- |
| A | 0.15 | 010 | 3 | 0.45 |
| B | 0.3 | 1 | 1 | 0.3 |
| C | 0.1 | 000 | 4 | 0.4 |
| D | 0.1 | 001 | 4 | 0.4 |
| E | 0.2 | 11 | 2 | 0.4 |
| F | 0.15 | 0010 | 3 | 0.45 |
将最后一列的值求和,得到WPL:
WPL = 0.45 + 0.3 + 0.4 + 0.4 + 0.4 + 0.45 = 2.4
因此,带权路径长度WPL为2.4。
3. 求A、B、C、D、E、F的Huffman编码。
根据Huffman树的构造过程,我们可以得到以下编码:
| 字母 | 频率 | 编码 |
| --- | --- | --- |
| A | 0.15 | 010 |
| B | 0.3 | 1 |
| C | 0.1 | 000 |
| D | 0.1 | 001 |
| E | 0.2 | 11 |
| F | 0.15 | 0010 |
因此,A的Huffman编码为010,B的Huffman编码为1,C的Huffman编码为000,D的Huffman编码为001,E的Huffman编码为11,F的Huffman编码为0010。
希望我的回答能够帮到你。
给定权值构造哈夫曼树C语言代码
以下是使用C语言实现以权值为基础构建哈夫曼树的示例代码:
```
#include <stdio.h>
#include <stdlib.h>
#define MAX_TREE_HT 100
struct MinHeapNode {
char data;
unsigned freq;
struct MinHeapNode *left, *right;
};
struct MinHeap {
unsigned size;
unsigned capacity;
struct MinHeapNode** array;
};
struct MinHeapNode* newNode(char data, unsigned freq) {
struct MinHeapNode* temp = (struct MinHeapNode*)malloc(sizeof(struct MinHeapNode));
temp->left = temp->right = NULL;
temp->data = data;
temp->freq = freq;
return temp;
}
struct MinHeap* createMinHeap(unsigned capacity) {
struct MinHeap* minHeap = (struct MinHeap*)malloc(sizeof(struct MinHeap));
minHeap->size = 0;
minHeap->capacity = capacity;
minHeap->array = (struct MinHeapNode**)malloc(minHeap->capacity * sizeof(struct MinHeapNode*));
return minHeap;
}
void swapMinHeapNode(struct MinHeapNode** a, struct MinHeapNode** b) {
struct MinHeapNode* t = *a;
*a = *b;
*b = t;
}
void minHeapify(struct MinHeap* minHeap, int idx) {
int smallest = idx;
int left = 2 * idx + 1;
int right = 2 * idx + 2;
if (left < minHeap->size && minHeap->array[left]->freq < minHeap->array[smallest]->freq)
smallest = left;
if (right < minHeap->size && minHeap->array[right]->freq < minHeap->array[smallest]->freq)
smallest = right;
if (smallest != idx) {
swapMinHeapNode(&minHeap->array[smallest], &minHeap->array[idx]);
minHeapify(minHeap, smallest);
}
}
int isSizeOne(struct MinHeap* minHeap) {
return (minHeap->size == 1);
}
struct MinHeapNode* extractMin(struct MinHeap* minHeap) {
struct MinHeapNode* temp = minHeap->array[0];
minHeap->array[0] = minHeap->array[minHeap->size - 1];
--minHeap->size;
minHeapify(minHeap, 0);
return temp;
}
void insertMinHeap(struct MinHeap* minHeap, struct MinHeapNode* minHeapNode) {
++minHeap->size;
int i = minHeap->size - 1;
while (i && minHeapNode->freq < minHeap->array[(i - 1) / 2]->freq) {
minHeap->array[i] = minHeap->array[(i - 1) / 2];
i = (i - 1) / 2;
}
minHeap->array[i] = minHeapNode;
}
void buildMinHeap(struct MinHeap* minHeap) {
int n = minHeap->size - 1;
int i;
for (i = (n - 1) / 2; i >= 0; --i)
minHeapify(minHeap, i);
}
void printArr(int arr[], int n) {
int i;
for (i = 0; i < n; ++i)
printf("%d", arr[i]);
printf("\n");
}
int isLeaf(struct MinHeapNode* root) {
return !(root->left) && !(root->right);
}
struct MinHeap* createAndBuildMinHeap(char data[], int freq[], int size) {
struct MinHeap* minHeap = createMinHeap(size);
for (int i = 0; i < size; ++i)
minHeap->array[i] = newNode(data[i], freq[i]);
minHeap->size = size;
buildMinHeap(minHeap);
return minHeap;
}
struct MinHeapNode* buildHuffmanTree(char data[], int freq[], int size) {
struct MinHeapNode *left, *right, *top;
struct MinHeap* minHeap = createAndBuildMinHeap(data, freq, size);
while (!isSizeOne(minHeap)) {
left = extractMin(minHeap);
right = extractMin(minHeap);
top = newNode('$', left->freq + right->freq);
top->left = left;
top->right = right;
insertMinHeap(minHeap, top);
}
return extractMin(minHeap);
}
void printCodes(struct MinHeapNode* root, int arr[], int top) {
if (root->left) {
arr[top] = 0;
printCodes(root->left, arr, top + 1);
}
if (root->right) {
arr[top] = 1;
printCodes(root->right, arr, top + 1);
}
if (isLeaf(root)) {
printf("%c: ", root->data);
printArr(arr, top);
}
}
void HuffmanCodes(char data[], int freq[], int size) {
struct MinHeapNode* root = buildHuffmanTree(data, freq, size);
int arr[MAX_TREE_HT], top = 0;
printCodes(root, arr, top);
}
int main() {
char arr[] = { 'a', 'b', 'c', 'd', 'e', 'f' };
int freq[] = { 5, 9, 12, 13, 16, 45 };
int size = sizeof(arr) / sizeof(arr[0]);
HuffmanCodes(arr, freq, size);
return 0;
}
```
这个示例代码使用了最小堆来实现哈夫曼树的构建。它首先创建一个最小堆,并将每个字符及其频率的节点插入堆中。然后它从堆中提取两个最小频率的节点并将它们合并为一个新节点,这个新节点的频率等于两个节点的频率之和。然后将这个新节点插入到堆中并重复此过程,直到堆中只剩下一个节点,即哈夫曼树的根节点。最后,它遍历哈夫曼树并输出每个字符的哈夫曼编码。