输入五个字符并根据它出现的概率生成一颗哈夫曼数

1、在VC++平台编制程序，根据输入的字符（最多为五个字符）及其出现的概率能够生成一棵Huffman树。

VC2003平台, 成功编译运行，测试结果正确

测试数据字符数5

测试码文='a','e','r','t','d'

测试码文出现次数=8,4,6,3,1

测试电文1="01011101111100011";

测试电文2="0101011010";

#include "stdafx.h"

#include <stdio.h>
#include <string.h>
#define N 50 //叶子结点数/
#define M 2*N-1 //树中结点总数/

typedef struct
{
char data[5]; //结点值/
int weight; //权重/
int parent; //双亲结点/
int lchild; //左孩子结点/
int rchild; //右孩子结点/
} HTNode;

typedef struct
{
char cd[N]; //存放哈夫曼码/
int start;
} HCode;

void CreateHT(HTNode ht[],int n)
{
int i,k,lnode,rnode;
int min1,min2;
for (i=0;i<2*n-1;i++) //所有结点的相关域置初值-1/
ht[i].parent=ht[i].lchild=ht[i].rchild=-1;
for (i=n;i<2*n-1;i++) //构造哈夫曼树/
{
min1=min2=32767; //lnode和rnode为最小权重的两个结点位置/
lnode=rnode=-1;
for (k=0;k<=i-1;k++)
if (ht[k].parent==-1) //只在尚未构造二叉树的结点中查找/
{
if (ht[k].weight<min1)
{
min2=min1;rnode=lnode;
min1=ht[k].weight;lnode=k;
}
else if (ht[k].weight<min2)
{
min2=ht[k].weight;rnode=k;
}
}
ht[lnode].parent=i;ht[rnode].parent=i;
ht[i].weight=ht[lnode].weight+ht[rnode].weight;
ht[i].lchild=lnode;ht[i].rchild=rnode;
}
}

void CreateHCode(HTNode ht[],HCode hcd[],int n)
{
int i,f,c;
HCode hc;
for (i=0;i<n;i++) //根据哈夫曼树求哈夫曼编码/
{
hc.start=n;c=i;
f=ht[i].parent;
while (f!=-1) //循序直到树根结点/
{
if (ht[f].lchild==c) //处理左孩子结点/
hc.cd[hc.start--]='0';
else //处理右孩子结点/
hc.cd[hc.start--]='1';
c=f;f=ht[f].parent;
}
hc.start++; //start指向哈夫曼编码最开始字符/
hcd[i]=hc;
}
}

void DispHCode(HTNode ht[],HCode hcd[],int n)
{
int i,k;
int sum=0,m=0;
printf(" 输出哈夫曼编码:\n"); //输出哈夫曼编码/
for (i=0;i<n;i++)
{
int j=0;
printf(" %s:\t",ht[i].data);
for (k=hcd[i].start;k<=n;k++)
{
printf("%c",hcd[i].cd[k]);
j++;
}
m+=ht[i].weight;
sum+=ht[i].weight*j;
printf("\n");
}
printf("\n 平均长度=%g\n",1.0*sum/m);
}
int findsubstr(char* S, int cS, HCode hcd[],int n, int &pos)
{
int i,k, j;
if (pos > cS) return -1;
for (i=0;i<n;i++)
{
j = pos;
for (k=hcd[i].start;k<=n;k++)
{
if (S[j] == hcd[i].cd[k])
j++;
else
break;
}
if (k > n)
{
pos = j;
return i;
}
}
return -1;
}

bool Recode(char* S, int cS, HTNode ht[],HCode hcd[],int n){
char strresult[80]={'\0'};
int pos = 0;
while (pos < cS)
{
int result = findsubstr(S, cS, hcd, n, pos);
if (result != -1)
{
strcat(strresult, ht[result].data);
}
else
return false;
}
printf("译码结果为：%s\n\n", strresult);
return true;
}
void main()
{
int i, n;
char str[6][2];
int fnum[6];
memset(str,'\0',12*sizeof(char));
printf("输入字符个数\n");
scanf("%d", &n);
getchar();

printf("开始录入字符\n");
for (int i = 0;i<n;i++)
{
scanf("%c",str[i]);
getchar();
}

printf("开始录入字符出现次数\n");

for (int i = 0;i<n;i++)
{
scanf("%d",&fnum[i]);
getchar();
}

HTNode ht[M];
HCode hcd[N];
for (i=0; i<n; i++)
{
strcpy(ht[i].data, str[i]);
ht[i].weight=fnum[i];
}
printf("\n");
CreateHT(ht,n);
CreateHCode(ht,hcd,n);
DispHCode(ht,hcd,n);
printf("\n");

char* strtest1="01011101111100011";
printf("破译原文是：%s\n", strtest1);
int clength = strlen(strtest1);
if (!Recode(strtest1, clength, ht, hcd, n))
printf("译密失败\n");

char* strtest2="0101011010";
printf("破译原文是：%s\n", strtest2);
clength = strlen(strtest2);
if (!Recode(strtest2, clength, ht, hcd, n))
printf("译密失败\n");

system("pause");
}