Huffman编码

  1 #define _CRT_SECURE_NO_WARNINGS
  2 #include <iostream>
  3 #include <cstdio>
  4 #include <cstring>
  5 #include <cstdlib>
  6 #include <string>
  7 using namespace std;
  8 
  9 struct Node {
 10     char character;
 11     unsigned int weight;
 12     unsigned int parent;
 13     unsigned int lchild;
 14     unsigned int rchild;
 15 };
 16 
 17 Node *HT = NULL;
 18 char **HC = NULL;
 19 int n = 0;
 20 
 21 void CreatHuffmanTree(void);
 22 void CodeHuffman(void);
 23 void Encode(void);
 24 void Decode(void);
 25 void Free(void);
 26 void Select(const int len, int &a, int &b);
 27 int Search(const char &key);
 28 
 29 int main(void)
 30 {
 31     freopen("cin.txt", "r", stdin);
 32     freopen("cout.txt", "w", stdout);
 33     while (cout << "请输入字符个数：
", cin >> n) {    //    n>=2
 34         getchar();                //    吸收回车，防止空格字符的输入异常
 35 
 36         CreatHuffmanTree();        //    建哈夫曼树
 37 
 38         CodeHuffman();            //    为每个字符编码
 39 
 40         Encode();                //    为明文编码
 41 
 42         Decode();                //    从密文译码
 43 
 44         Free();                    //    释放动态分配的空间
 45     }
 46     system("pause");// 有用？
 47     return 0;
 48 }
 49 
 50 void CreatHuffmanTree(void)
 51 {
 52     // weight存放n个字符的权值(均>0)，构造哈夫曼树HT，
 53     // 并求出n个字符的哈夫曼编码HC
 54     if (n <= 1)
 55         return;
 56     int m = 2 * n - 1;
 57 
 58     HT = (Node *)malloc((m + 1) * sizeof(Node));    //    0号单元未用
 59     if( HT == NULL) {
 60         cerr << "Allocate failed!" << endl;
 61         exit(OVERFLOW);
 62     }
 63     memset(HT, 0, (m + 1) * sizeof(Node));
 64 
 65     cout << "请输入各字符及其权值：" << endl;
 66     for (int i = 1; i <= n; i++) {
 67         HT[i].character = getchar();
 68         cin >> HT[i].weight;                        //    输入字符及权值的信息
 69         getchar();
 70     }
 71     cout << n << "个字符及其权值：" << endl;
 72     for (int i = 1; i <= n; i++) {
 73         printf("%c(%3d) ", HT[i].character, HT[i].weight);
 74         if (i % 10 == 0)
 75             cout << endl;
 76     }
 77     cout << endl;
 78 
 79     /*    printf("
哈夫曼树的构造过程如下所示：
");
 80     printf("HT初态:
  结点  weight  parent  lchild  rchild");
 81     for (int i = 1; i <= m; i++)
 82     printf("
%4d%8d%8d%8d%8d", i, HT[i].weight, HT[i].parent, HT[i].lchild, HT[i].rchild);
 83     printf("    按回车键，继续 ...");
 84     //    getchar();*/
 85 
 86     for (int i = n + 1; i <= m; i++) {
 87         // 在HT[1..i-1]中选择parent为0且weight最小的两个结点，
 88         // 其序号分别为s1和s2。
 89         int s1(0);
 90         int s2(0);
 91         Select(i - 1, s1, s2);
 92         HT[s1].parent = i;
 93         HT[s2].parent = i;
 94         HT[i].lchild = s1;
 95         HT[i].rchild = s2;
 96         HT[i].weight = HT[s1].weight + HT[s2].weight;
 97 
 98         /*        printf("
select: s1=%d   s2=%d

", s1, s2);
 99         printf("  结点  weight  parent  lchild  rchild");
100         for (int j = 1; j <= i; j++)
101         printf("
%4d%8d%8d%8d%8d", j, HT[j].weight, HT[j].parent, HT[j].lchild, HT[j].rchild);
102         printf("    按回车键，继续 ...");
103         //        getchar();*/
104     }
105 }
106 
107 void CodeHuffman(void)
108 {
109     //--- 从叶子到根逆向求每个字符的哈夫曼编码 ---
110     HC = (char **)malloc((n + 1) * sizeof(char *));            //    指针数组
111     if (HC == NULL) {
112         cerr << "Allocate failed!" << endl;
113         exit(OVERFLOW);
114     }
115     memset(HC, 0, (n + 1) * sizeof(char *));                //    0号单元未用
116 
117     char *cd = (char *)malloc(n * sizeof(char));            // 分配求编码的工作空间
118     if (cd == NULL) {
119         cerr << "Allocate failed!" << endl;
120         exit(OVERFLOW);
121     }
122     memset(cd, 0, n * sizeof(char));                        // 编码结束符。
123 
124     for (int i = 1; i <= n; ++i) {                            // 逐个字符求哈夫曼编码
125         int start = n - 1;                                    // 编码结束符位置
126         unsigned int c(0);
127         unsigned int f(0);
128         for (c = i, f = HT[i].parent; f != 0; c = f, f = HT[f].parent) {                                                        
129             if (HT[f].lchild == c)                            // 从叶子到根逆向求编码
130                 cd[--start] = '0';
131             else
132                 cd[--start] = '1';            
133         }
134 
135         HC[i] = (char *)malloc((n - start) * sizeof(char));    // 为第i个字符编码分配空间
136         if (HC[i] == NULL) {
137             cerr << "Allocate failed!" << endl;
138             exit(OVERFLOW);
139         }
140 
141         strcpy(HC[i], &cd[start]);                            // 从cd复制编码(串)到HC
142     }
143     free(cd);
144     cd = NULL;
145 
146     printf("

各字符的哈夫曼编码如下：
");
147     for(int i = 1; i <= n; i++) {
148         printf("%c(%3d): %s
" ,HT[i].character, HT[i].weight, HC[i]);
149     }
150 }
151 
152 void Encode(void)
153 {
154     cout << "请输入报文：" << endl;
155     string str;
156     getline(cin, str);
157     for (unsigned int i = 0; i < str.size(); i++) {
158         int loc = Search(str[i]);
159         cout << HC[loc] << endl;
160     }
161     cout << endl;
162 }
163 
164 void Decode(void)
165 {
166     cout << "请输入密文：" << endl;
167     string str;
168     cin >> str;
169     for (unsigned int i = 0; i < str.size();) {
170         int cursor = 2 * n - 1;
171         while (HT[cursor].lchild != 0 && HT[cursor].rchild != 0) {    //    全非0
172             switch (str[i++]) {
173 
174             case '0':
175                 cursor = HT[cursor].lchild;
176                 break;
177 
178             case '1':
179                 cursor = HT[cursor].rchild;
180                 break;
181 
182             default:
183                 cerr << "Input error!" << endl;
184                 exit (1);
185             }
186         }
187         if (!(HT[cursor].lchild == 0 && HT[cursor].rchild == 0)) {    //    此时应是全0
188             cerr << "Input error!" << endl;
189             exit (1);            
190         }
191         cout << HT[cursor].character << endl;
192     }
193 }
194 
195 void Free(void)
196 {
197     free(HT);    //    释放动态分配的空间
198     HT = NULL;
199     for(int i = 1; i <= n; i++) {
200         free(HC[i]);
201         HC[i] = NULL;
202     }
203     free(HC);
204     HC = NULL;
205 }
206 
207 void Select(const int len, int &a, int &b)
208 {
209     // 在HT[1..len]中选择parent为0且weight最小的两个结点，
210     // 其序号分别为a和b。
211     for (int i = 1; i <= len; i++) {
212         if (HT[i].parent == 0) {
213             a = i;
214             break;
215         }
216     }
217     for (int i = a + 1; i <= len; i++) {
218         if (HT[i].parent == 0) {
219             b = i;
220             break;
221         }
222     }
223     if (HT[a].weight > HT[b].weight)    //    保持a的始终小于b的，有序
224         swap(a, b);
225     for (int i = max(a, b) + 1; i <= len; i++) {
226         if (HT[i].parent == 0) {
227             if (HT[i].weight > HT[b].weight)
228                 continue;
229             else if (HT[i].weight < HT[a].weight) {
230                 b = a;
231                 a = i;
232             }
233             else
234                 b = i;
235         }
236     }
237     if ( !(HT[a].parent == 0 && HT[b].parent == 0) ) {
238         cerr << "Error!" << endl;
239         exit(1);
240     }
241 }
242 
243 int Search(const char &key)        //    找到key在HT中的下标，找不到则返回0
244 {
245     HT[0].character = key;        //    哨兵,顺序查找
246     int i;
247     for (i = n; HT[i].character != key; --i);
248     if (i == 0) {
249         cerr << "Inexistence!" << endl;
250         exit(1);
251     }
252     return i;
253 }
254 
255 /*
256 cin.txt:
257 27
258   186
259 a 64
260 b 13
261 c 22
262 d 32
263 e 103
264 f 21
265 g 15
266 h 47
267 i 57
268 j 1
269 k 5
270 l 32
271 m 20
272 n 57
273 o 63
274 p 15
275 q 1
276 r 48
277 s 51
278 t 80
279 u 23
280 v 8
281 w 18
282 x 1
283 y 16
284 z 1
285 this program is my favorite
286 */

 输出：

请输入字符个数：
请输入各字符及其权值：
27个字符及其权值：
(186) a( 64) b( 13) c( 22) d( 32) e(103) f( 21) g( 15) h( 47) i( 57)
j( 1) k( 5) l( 32) m( 20) n( 57) o( 63) p( 15) q( 1) r( 48) s( 51)
t( 80) u( 23) v( 8) w( 18) x( 1) y( 16) z( 1)


各字符的哈夫曼编码如下：
(186): 111
a( 64): 1010
b( 13): 100000
c( 22): 00000
d( 32): 10110
e(103): 010
f( 21): 110011
g( 15): 100010
h( 47): 0001
i( 57): 0110
j( 1): 1100001000
k( 5): 11000011
l( 32): 10111
m( 20): 110010
n( 57): 0111
o( 63): 1001
p( 15): 100001
q( 1): 1100001010
r( 48): 0010
s( 51): 0011
t( 80): 1101
u( 23): 00001
v( 8): 1100000
w( 18): 110001
x( 1): 1100001011
y( 16): 100011
z( 1): 1100001001
请输入报文：
1101
0001
0110
0011
111
100001
0010
1001
100010
0010
1010
110010
111
0110
0011
111
110010
100011
111
110011
1010
1100000
1001
0010
0110
1101
010

请输入密文：
请按任意键继续. . .
请输入字符个数：

考研机试题

  1 /**************************************
  2 created:    2014/03/14
  3 filename:    13_07_huffman.cpp
  4 purpose:    Huffman encode
  5 **************************************/
  6 
  7 #pragma warning(disable: 4786)
  8 #include <iostream>
  9 #include <queue>
 10 #include <string>
 11 #include <cmath>
 12 #include <map>
 13 #include <cassert>
 14 using namespace std;
 15 
 16 struct Node {
 17     double weight;
 18     Node *left;
 19     Node *right;
 20     int num;    // 记录输入时的次序，为了按输入顺序输出
 21     
 22     Node(double w = 0.0, int n = -1) : weight(w), left(NULL), right(NULL), num(n) {}
 23     
 24     bool is_leaf() const {
 25         return NULL == left && NULL == right;
 26     }
 27 };
 28 
 29 struct Cmp {
 30     bool operator()(Node *a, Node *b) const {
 31         return a->weight > b->weight;
 32     }
 33 };
 34 
 35 /*
 36 ** 使两个结点成为兄弟，返回他们的父亲的地址
 37 */
 38 Node * conjoin(Node *left, Node *right) {
 39     Node *father = new Node(left->weight + right->weight);    // 这里产生的结点都是分支结点
 40     father->left = left;
 41     father->right = right;
 42     
 43     return father;
 44 }
 45 
 46 /*
 47 ** 根据优先队列建 Huffman 树
 48 */
 49 Node * create_tree(priority_queue<Node *, vector<Node *>, Cmp> &Q) {
 50     while (1 != Q.size()) {
 51         Node *left = Q.top();
 52         Q.pop();
 53         Node *right = Q.top();
 54         Q.pop();
 55         Q.push(conjoin(left, right));
 56     }
 57     Node *tree = Q.top();
 58     Q.pop();
 59     
 60     return tree;
 61 }
 62 
 63 /*
 64 ** 前序遍历 Huffman 树得到叶子的编码存储在 code_map 中
 65 */
 66 void get_code(Node const *const tree, const string str, map<int, string> &code_map) {
 67     if (NULL == tree) {
 68         return ;
 69     }
 70     if (tree->is_leaf()) {
 71         code_map[tree->num] = str;
 72         return ;
 73     }
 74     get_code(tree->left, str + "0", code_map);
 75     get_code(tree->right, str + "1", code_map);
 76 }
 77 
 78 /*
 79 ** 后序遍历释放树
 80 */
 81 void free_tree(Node *&tree) {
 82     if (NULL == tree) {
 83         return ;
 84     }
 85     
 86     free(tree->left);
 87     free(tree->right);
 88     delete tree;
 89     tree = NULL;
 90 }
 91 
 92 int main(int argc, char **argv) {
 93     FILE *in_stream = freopen("7.in", "r", stdin);
 94     if (NULL == in_stream) {
 95         cerr << "Can not open file." << endl;
 96         exit(1);
 97     }
 98     FILE *out_stream = freopen("7.out", "w", stdout);
 99     
100     priority_queue<Node *, vector<Node *>, Cmp> Q;
101     double elem;
102     int n;
103     cin >> n;
104     int i;
105     for (i = 0; i < n; ++i) {
106         cin >> elem;
107         Q.push(new Node(elem, i));
108     }
109     
110     Node *tree = create_tree(Q);        // 根据输入数据建 Huffman 树
111     assert(Q.empty());
112     map<int, string> code_map;
113     get_code(tree, string(), code_map);    // 根据 Huffman 树获取 Huffman 编码
114     
115     for (i = 0; i < n; ++i) {
116         cout << code_map[i] << endl;
117     }
118     
119     free_tree(tree);
120     
121     fclose(in_stream);
122     fclose(out_stream);
123     
124     return 0;
125 }
126 
127 /* 7.in
128 4
129 0.3 0.1 0.4 0.2
130 */
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原文地址：https://www.cnblogs.com/jjtx/p/3643964.html