数据结构实验之二叉树六：哈夫曼编码

Problem Description

字符的编码方式有多种，除了大家熟悉的ASCII编码，哈夫曼编码(Huffman Coding)也是一种编码方式，它是可变字长编码。该方法完全依据字符出现概率来构造出平均长度最短的编码，称之为最优编码。哈夫曼编码常被用于数据文件压缩中，其压缩率通常在20%～90%之间。你的任务是对从键盘输入的一个字符串求出它的ASCII编码长度和哈夫曼编码长度的比值。

Input

 输入数据有多组，每组数据一行，表示要编码的字符串。

Output

 对应字符的ASCII编码长度la，huffman编码长度lh和la/lh的值(保留一位小数)，数据之间以空格间隔。

Example Input

AAAAABCDTHE_CAT_IN_THE_HAT

Example Output

64 13 4.9144 51 2.8

#include <stdio.h>#include <stdlib.h>#include <string.h>#define SUM 10typedef int Elemtype;typedef struct BiTnode{    Elemtype data;    struct BiTnode *lchild, *rchild, *next, *parent;} BiTnode, *BiTree;typedef struct{    BiTree front;    BiTree rear;    int len;} Queue;Queue q;int wl;void Sort(int a[], int n);void Init();void EnQueue(int x);BiTree OutQueue();void Insert(BiTree root);BiTree Create();int Path(BiTree p);void WPL(BiTree p);int main(){    int n, i, length;    char s[10010];    while(~scanf("%s", s))    {        n = 0;        wl = 0;        int la;        int b[150] = {0};        int a[150];        length = strlen(s);        la = 8 * length;        for(i = 0; i < length; i++)            b[s[i]]++;        for(i = 0; i < 150; i++)            if(b[i] != 0)                a[n++] = b[i];        Sort(a, n);        Init();        for(i = 0;i < n; i++)            EnQueue(a[i]);        BiTree root;        root = Create();        WPL(root);        printf("%d %d %.1lf\n", la, wl, 1.0*la/wl);    }    return 0;}void Sort(int a[], int n){    int i, j;    for(i = 0; i < n-1; i++)        for(j = 0; j < n - i - 1; j++)        {            if(a[j] > a[j+1])            {                int t = a[j];                a[j] = a[j+1];                a[j+1] = t;            }        }}void Init(){    q.front = (BiTree)malloc(sizeof(BiTnode));    q.front -> next = NULL;    q.rear = q.front;    q.len = 0;}void EnQueue(int x){    BiTree p;    p = (BiTree)malloc(sizeof(BiTnode));    p -> next = NULL;    p -> rchild = NULL;    p -> lchild = NULL;    p -> parent = NULL;    p -> data = x;    q.rear -> next = p;    q.rear = p;    q.len++;}BiTree OutQueue(){    if(q.len == 1)    {        BiTree p;        p = q.front -> next;        q.front = NULL;        q.rear = q.front;        q.len--;        return p;    }    else    {        BiTree p, t;        t = p = q.front -> next;        p = p -> next;        q.front -> next = p;        q.len--;        return t;    }}void Insert(BiTree root){    BiTree p, t;    p = q.front -> next;    t = q.front;    while(root -> data > p -> data)    {        p = p -> next;        t = t -> next;        if(!p)            break;    }    if(p)    {        root -> next = p;        t -> next = root;    }    else    {        q.rear -> next = root;        q.rear = root;    }    q.len++;}BiTree Create(){    if(q.len > 1)    {        BiTree root, t1, t2;        root=(BiTree)malloc(sizeof(BiTnode));        root -> next = NULL;        root -> parent = NULL;        t1 = OutQueue();        t2 = OutQueue();        root -> lchild = t1;        root -> rchild = t2;        root -> data = t1 -> data + t2 -> data;        t1 -> parent = root;        t2 -> parent = root;        if(q.len == 0)            return root;        else        {            Insert(root);            Create();        }    }}int Path(BiTree p){    int l = 0;    int x = p -> data;    while(p -> parent)    {        p = p -> parent;        l++;    }    return l * x;}void WPL(BiTree p){    if(!p -> lchild && !p -> rchild)    {        int e = Path(p);        wl += e;    }    else    {        if(p -> lchild)            WPL(p -> lchild);        if(p -> rchild)            WPL(p -> rchild);    }}