哈夫曼树
#include <stdio.h>
#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; /*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;
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[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;
double sum=0,m=0;
int j;
printf("输出哈夫曼编码:\n"); /*输出哈夫曼编码*/
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");
}
}
void main()
{
int n=5,i; /*n表示初始字符串的个数*/
char str[]={'a','b','c','d','e'};
double fnum[]={4,2,1,7,3};
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");
}