哈夫曼二叉樹原始碼 (轉)

哈夫曼二叉樹原始碼 (轉)[@more@]

哈夫曼二叉樹原始碼：

給定一個字串，根據統計字串中各個字元出現的頻率對字元進行哈夫曼編碼，然後對原字串進行編碼，並輸出編碼後的內容

——資料結構

#include :namespace prefix = o ns = "urn:schemas--com::office" />

 

#define MAX_NODE 1024

#define MAX_WEIGHT  4096

 

typedef struct HaffmanTreeNode {

  char ch, code[15];

  int weight;

  int parent, lchild, rchild;

} HTNode, *HaTree;

 

typedef struct {

  HTNode arr[MAX_NODE];

  int total;

} HTree;

 

/*  統計字元出現的頻率  */

int statistic_char(char *text, HTree *t){

  int i, j;

  int text_len = strlen(text);

  t->total = 0;

  for (i=0; i

  for (j=0; jtotal; j++) if (t->arr[j].ch == text[i]){

    t->arr[j].weight ++;

    break;

  }

  if (j==t->total) {

    t->arr[t->total].ch = text[i];

    t->arr[t->total].weight = 1;

    t->total ++;

  }

  }

  printf("char frequence ");

  for (i=0; itotal; i++)

    printf("'%c'  %d ", t->arr[i].ch, t->arr[i].weight);

  printf(" ");

  return t->total;

}

 

int create_htree(HTree *t)

{

  int i;

  int total_node = t->total * 2 - 1;

  int min1, min2; /* 權最小的兩個結點 */

  int min1_i, min2_i; /*  權最小結點對應的編號  */

  int leaves = t->total;

  for (i=0; i

    t->arr[i].parent = -1;

    t->arr[i].rchild = -1;

    t->arr[i].lchild = -1;

  }

  while (t->total < total_node) {

  min1 = min2 = MAX_WEIGHT;

  for (i=0; itotal; i++) { /*  對每一個結點  */

    if (t->arr[i].parent == -1 /*  結點沒有被合併  */

    && t->arr[i].weight < min2) { /*  結點的權比最小權小  */

    if (t->arr[i].weight < min1) { /*  如果它比最小的結點還小  */

      min2_i = min1_i;  min2 = min1;

      min1_i = i;  min1 = t->arr[i].weight;

    }

    else

    {

      min2_i = i;  min2 = t->arr[i].weight;

    }

  }

  }

    t->arr[t->total].weight = min1 + min2;

    t->arr[t->total].parent = -1;

    t->arr[t->total].lchild = min1_i;

    t->arr[t->total].rchild = min2_i;

    t->arr[min1_i].parent = t->total;

    t->arr[min2_i].parent = t->total;

    t->arr[t->total].ch = ' ';

    t->total ++;

  }

  return 0;

}

 

/*  對哈夫曼樹進行編碼  */

void coding(HTree *t, int head_i, char *code)

{

  if ( head_i == -1) return;

  if (t->arr[head_i].lchild == -1 && t->arr[head_i].rchild == -1) {

    strcpy(t->arr[head_i].code, code);

    printf("'%c': %s ", t->arr[head_i].ch, t->arr[head_i].code);

  }

  else {

  int len = strlen(code);

    strcat(code, "0");

    coding(t, t->arr[head_i].lchild, code);

    code[len] = '1';

    coding(t, t->arr[head_i].rchild, code);

    code[len] = '';

  }

}

 

/*  中序列印樹  */

void print_htree_ldr(HTree *t, int head_i, int deep, int* path)

{

  int i;

  if (head_i == -1) return;

  path[deep] = 0;

  print_htree_ldr(t, t->arr[head_i].lchild, deep + 1, path);

  if (deep) printf("  ");

  for (i=1; i

  int dir = path[i]==path[i-1];

  if ( t->arr[head_i].lchild == -1 && t->arr[head_i].rchild == -1)

    printf("%s── %d '%c' ", dir? "┌":"└",

    t->arr[head_i].weight, t->arr[head_i].ch);

  else if (head_i != t->total-1)

    printf("%s─%02d┤ ", dir? "┌":"└"", t->arr[head_i].weight);

  else

    printf("  %02d┤ ", t->arr[head_i].weight);

  path[deep] = 1;

  print_htree_ldr(t, t->arr[head_i].rchild, deep + 1, path);

}

 

/*  對字元進行編碼  */

void code_string(char *text, HTree *t)

{

  int i, j, text_len = strlen(text);

  int n = 0;

  for (i=0; i

  char ch = text[i];

  for (j=0; jtotal; j++) if (ch == t->arr[j].ch) {

    printf("%s ", t->arr[j].code);

  n += strlen(t->arr[j].code);

    break;

  }

  }

  printf(" %d chars, Total len = %d ", text_len, n);

}

 

int main(int argc, char* argv[])

{

  HTree t;

  char text[128]="ABAAAAEEEAAACCCCAAAACCDEA";

  char code[128] = "";

  int path[16]={0};

  statistic_char(text, &t);

  create_htree(&t);

  print_htree_ldr(&t, t.total-1, 0, path);

  coding(&t, t.total-1, code);

  code_string(text, &t);

  return 0;

}

 

 

輸出結果：

char frequence
'A'  13
'B'  1
'E'  4
'C'  6
'D'  1

  ┌── 6 'C'
  ┌─12┤
  │  │  ┌── 1 'B'
  │  │  ┌─02┤
  │  │  │  └── 1 'D'
  │  └─06┤
  │  └── 4 'E'
  25┤
  └── 13 'A'
'C': 00
'B': 0100
'D': 0101
'E': 011
'A': 1
1 0100 1 1 1 1 011 011 011 1 1 1 00 00 00 00 1 1 1 1 00 00 0101 011 1