请帮忙解释个程序（哈夫曼编码）

[APIandMFC]

#include<stdio.h>
#include<stdlib.h>
#include<string.h>
typedef struct
{
	unsigned int weight;
	unsigned int parent,lchild,rchild;
}HTNode,*HuffmanTree;
typedef char **HuffmanCode;
int min(HuffmanTree t,int i)
{
	int j,flag;
	unsigned int k=1000000;//k不小于可能的值
	for(j=1;j<=i;j++)
		if(t[j].weight<k && t[j].parent==0)
		{
			k=t[j].weight;
			flag=j;
		}
	t[flag].parent=1;
	return flag;
}
void select(HuffmanTree t,int i,int &s1,int &s2)
{
	int j;
	s1=min(t,i);
	s2=min(t,i);
	if(s1>s2)
	{
		j=s1;
		s1=s2;
		s2=j;
	}
}
void HuffmanCoding(HuffmanTree &HT,HuffmanCode &HC,int *w,int n)
{
	//w存放n个字符的权值（均>0），构造哈夫曼树HT，并求出n个字符的哈夫曼编码HC
	int m,i,s1,s2;
	unsigned c,f;
	HuffmanTree p;
	char *cd;
	if(n<=1)
		return;
	m=2*n-1;
	HT=(HuffmanTree)malloc((m+1)*sizeof(HTNode));//0号单元未用
	for(p=HT+1,i=1;i<=n;++i,++p,++w)//初始化每个结点
	{
		(*p).weight=*w;
		(*p).parent=0;
		(*p).lchild=0;
		(*p).rchild=0;
	}
	for(;i<=m;++i,++p)
		(*p).parent=0;
	for(i=n+1;i<=m;++i)//建哈夫曼树
	{
		//在HT[1~i-1]中选择parent为0且weight最小的两个结点，其序号分别为s1和s2
		select(HT,i-1,s1,s2);
		HT[s1].parent=HT[s2].parent=i;
		HT[i].lchild=s1;
		HT[i].rchild=s2;
		HT[i].weight=HT[s1].weight+HT[s2].weight;
	}
	//从叶子到根逆向求每个字符的哈夫曼编码
	HC=(HuffmanCode)malloc((n+1)*sizeof(char*));
	cd=(char*)malloc(n*sizeof(char));
	cd[n-1]='\0';
	int start;
	for(i=1;i<=n;i++)//解释一下这个循环
	{
		start=n-1;
		for(c=i,f=HT[i].parent;f!=0;c=f,f=HT[f].parent)
		{
			if(HT[f].lchild==c)
				cd[--start]='0';
			else
				cd[--start]='1';
		}
		HC[i]=(char*)malloc((n-start)*sizeof(char));
		strcpy(HC[i],&cd[start]);
	}
	free(cd);
}
void main()
{
	HuffmanTree HT;
	HuffmanCode HC;
	int *w,n,i;
	printf("请输入权值个数:");
	scanf("%d",&n);
	w=(int*)malloc(n*sizeof(int));
	printf("请一次输入%d个权值\n",n);
	for(i=0;i<=n-1;i++)
		scanf("%d",w+i);
	HuffmanCoding(HT,HC,w,n);
	for(i=1;i<=n;i++)
		puts(HC[i]);
}

解释一下 “从叶子到根逆向求每个字符的哈夫曼编码”那段代码。

[bfhtian]

帮顶

[lzpsweet]

Up