哈夫曼C语言代码-抄录

#include <stdio.h>
#define MAXBIT 100
#define MAXVALUE 10000  //
#define MAXLEAF 30 //叶节点数目
#define MAXNODE MAXLEAF*2-1 //节点数目
//首先构造出一颗哈夫曼树的结点,包括
//权重weight,父亲parent, 左儿子lchild,右儿子rchild。
typedef struct
{
int bit[MAXBIT];
int start;
}HCodeType;  //编码结构体
typedef struct HNodeType
{
int weight;
int parent;
int lchild;
int rchild;
}HNodeType; //结点结构体
//开始创建一颗哈夫曼树
void Huffman(HNodeType HuffNode[MAXNODE],int n)
{
int i,j,m1,m2,x1,x2; //i,j循环变量,m1,m2:构造哈弗曼树不同过程中两个最小权值结点的权值
//x1,x2 构造哈弗曼树不同过程中两个最小权值结点在数组中的序号
for(i=0;i<2*n-1;i++)//初始化哈弗曼树数组中的结点
{
HuffNode[i].weight = 0;
HuffNode[i].parent = -1;
HuffNode[i].lchild = -1;
HuffNode[i].rchild = -1;
}
//输入n个叶子结点的权值
for(i=0;i<n;i++)
{
printf("Plese input weight of leaf node %d:\n",i);
scanf("%d",&HuffNode[i].weight);
}
//循环构造哈弗曼树
for(i=0;i<n-1;i++)
{
m1=m2=MAXVALUE;//m1,m2中存放两个无父结点且结点权值最小的两个结点
x1=x2=0;
//找出所有结点中权值最小、无父节点的两个结点,并合并为一颗二叉树
for(j=0;j<n+i;j++)
{
if(HuffNode[j].weight<m1 && HuffNode[j].parent==-1)
{
m2 = m1;
x2 = x1;
m1 = HuffNode[j].weight;
x1 = j;
}
else if(HuffNode[j].weight < m2 && HuffNode[j].parent==-1)
{
m2 = HuffNode[j].weight;
x2 = j;
}
}
HuffNode[x1].parent = n+i;
HuffNode[x2].parent = n+i;
HuffNode[n+i].weight = HuffNode[x2].weight+HuffNode[x1].weight;
        HuffNode[n+i].lchild = x1;
HuffNode[n+i].rchild = x2;
}
}
int main()
{
HNodeType HuffNode[MAXNODE]; //定义一个结点结构体数组
HCodeType HuffCode[MAXLEAF],cd;//定义一个编码结构体数组,同时定义一个临时变量来存放求解编码时的信息
int i,j,c,n,p;
printf("Please input n:\n");
scanf("%d",&n);
Huffman(HuffNode,n);


for(i=0;i<n;i++)
{
cd.start = n-1;
c = i;
p = HuffNode[c].parent;
while(p!=-1) //父节点存在
{
if(HuffNode[p].lchild==c)
{
cd.bit[cd.start] = 0;
}
else
{
cd.bit[cd.start] = 1;
}
cd.start--;//求编码的低一位
c=p;
p=HuffNode[c].parent;//设置下一循环条件
}
for(j=cd.start+1;j<n;j++)
{
HuffCode[i].bit[j] = cd.bit[j];
}
HuffCode[i].start = cd.start;
}
for(i=0;i<n;i++)
{
printf("%d 's Huffman code is:",i);
for(j=HuffCode[i].start+1;j<n;j++)
{
printf("%d",HuffCode[i].bit[j]);
}
printf("\n");
}
getchar();
return 0;
}
posted @ 2016-10-27 18:15  Philtell  阅读(230)  评论(0编辑  收藏  举报