qq_25005337 于 2015.06.30 09:53 提问

3个回答

lzp_lrp      2015.06.30 10:06

lzp_lrp      2015.06.30 10:05
NeQhk   2015.07.06 21:30

#include
#define N 50 /*叶子结点数*/
#define M 2*N-1 /*树中结点总数*/
typedef struct
{
char data; /*结点值*/
double weight; /*权重*/
int parent; /*双亲结点

/
int lchild; /

int rchild; /*右孩子结

} HTNode;
typedef struct
{
char cd[N]; /*存放哈夫

int start;
} HCode;
void CreateHT(HTNode ht[],int n)
{
int i,k,lnode,rnode;
double 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
``````

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

{
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[i].weight=ht

[lnode].weight+ht[rnode].weight;
ht[i].lchild=lnode;ht

[i].rchild=rnode;
ht[lnode].parent=i;
ht[rnode].parent=i;
}
}
void CreateHCode(HTNode ht[],HCode hcd

[],int n)
{
int i,f,c;
HCode hc;
for (i=0;i<n;i++) /*根据哈夫

{
hc.start=n;c=i;
f=ht[i].parent;
while (f!=-1) /*循序直到

{
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;
double sum=0,m=0;
int j;
printf("输出哈夫曼编码:\n"); /*输出

for (i=0;i<n;i++)
{
j=0;
printf(" %c:",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");
}
}
int main()
{
int n=5,i; /*n表示初始

char str[]={'C','A','S','T','B'};
double fnum[]={2,4,2,3,3};
HTNode ht[M];
HCode hcd[N];
for (i=0;i<n;i++)
{
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");
return 0;
}