数据结构(C语言)-哈夫曼编码求助!!

2025-04-07 14:20:18
推荐回答(2个)
回答1:

#include
#include
#include

typedef char* HuffmanCode;/*动态分配数组,存储哈夫曼编码*/

typedef struct
{
unsigned int weight ; /* 用来存放各个结点的权值*/
unsigned int parent, LChild,RChild ; /*指向双亲、孩子结点的指针*/
}HTNode, * HuffmanTree; /*动态分配数组,存储哈夫曼树*/

void select(HuffmanTree *ht,int n, int *s1, int *s2)
{
int i;
int min;
for(i=1; i<=n; i++)
{
if((*ht)[i].parent == 0)
{
min = i;
i = n+1;
}
}
for(i=1; i<=n; i++)
{
if((*ht)[i].parent == 0)
{
if((*ht)[i].weight < (*ht)[min].weight)
min = i;
}
}
*s1 = min;
for(i=1; i<=n; i++)
{
if((*ht)[i].parent == 0 && i!=(*s1))
{
min = i;
i = n+1;
}
}
for(i=1; i<=n; i++)
{
if((*ht)[i].parent == 0 && i!=(*s1))
{
if((*ht)[i].weight < (*ht)[min].weight)
min = i;
}
}
*s2 = min;
}

void CrtHuffmanTree(HuffmanTree *ht , int *w, int n)
{ /* w存放已知的n个权值,构造哈夫曼树ht */
int m,i;
int s1,s2;
m=2*n-1;
*ht=(HuffmanTree)malloc((m+1)*sizeof(HTNode)); /*0号单元未使用*/
for(i=1;i<=n;i++)
{/*1-n号放叶子结点,初始化*/
(*ht)[i].weight = w[i];
(*ht)[i].LChild = 0;
(*ht)[i].parent = 0;
(*ht)[i].RChild = 0;
}
for(i=n+1;i<=m;i++)
{
(*ht)[i].weight = 0;
(*ht)[i].LChild = 0;
(*ht)[i].parent = 0;
(*ht)[i].RChild = 0;
} /*非叶子结点初始化*/
/* ------------初始化完毕!对应算法步骤1---------*/

for(i=n+1;i<=m;i++) /*创建非叶子结点,建哈夫曼树*/
{ /*在(*ht)[1]~(*ht)[i-1]的范围内选择两个parent为0且weight最小的结点,其序号分别赋值给s1、s2返回*/
select(ht,i-1,&s1,&s2);
(*ht)[s1].parent=i;
(*ht)[s2].parent=i;
(*ht)[i].LChild=s1;
(*ht)[i].RChild=s2;
(*ht)[i].weight=(*ht)[s1].weight+(*ht)[s2].weight;
}
}/*哈夫曼树建立完毕*/
void outputHuffman(HuffmanTree HT, int m)
{
if(m!=0)
{
printf("%d ", HT[m].weight);
outputHuffman(HT,HT[m].LChild);
outputHuffman(HT,HT[m].RChild);
}
}

void CrtHuffmanCode(HuffmanTree *ht, HuffmanCode *hc, int n)
/*从叶子结点到根,逆向求每个叶子结点对应的哈夫曼编码*/
{
char *cd;
int i;
unsigned int c;
int start;
int p;
hc=(HuffmanCode *)malloc((n+1)*sizeof(char *)); /*分配n个编码的头指针*/
cd=(char * )malloc(n * sizeof(char )); /*分配求当前编码的工作空间*/
cd[n-1]='\0'; /*从右向左逐位存放编码,首先存放编码结束符*/
for(i=1;i<=n;i++) /*求n个叶子结点对应的哈夫曼编码*/
{
start=n-1; /*初始化编码起始指针*/
for(c=i,p=(*ht)[i].parent; p!=0; c=p,p=(*ht)[p].parent) /*从叶子到根结点求编码*/
if( (*ht)[p].LChild == c)
cd[--start]='0'; /*左分支标0*/
else
cd[--start]='1'; /*右分支标1*/
hc[i]=(char *)malloc((n-start)*sizeof(char)); /*为第i个编码分配空间*/
strcpy(hc[i],&cd[start]);
}
free(cd);
for(i=1;i<=n;i++)
printf("%d编码为%s\n",(*ht)[i].weight,hc[i]);
}

void main()
{
HuffmanTree HT;
HuffmanCode HC;
int *w;
int i,n; // the number of elements;
int wei; // the weight of a element;
int m;

printf("input the total number of the Huffman Tree:" );
scanf("%d",&n);
w=(int *)malloc((n+1)*sizeof(int));
for(i=1;i<=n;i++)
{
printf("input the %d element's weight:",i);
fflush(stdin);
scanf("%d",&wei);
w[i]=wei;
}

CrtHuffmanTree(&HT,w,n);
m = 2*n-1;
outputHuffman(HT,m);
printf("\n");
CrtHuffmanCode(&HT,&HC,n);

}

回答2:

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