数据结构上机【构造哈夫曼树，并输出对应节点的哈夫曼编码】

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
typedef struct
{
	int weight;
	int parent,lchild,rchild;
}HTNode,*HuffmanTree;
typedef char **HuffmanCode;
void Select(HuffmanTree HT,int len,int &s1,int &s2)
{//此函数找出所有结点中最小的两个 
	int i;
	int weight=1000000;
	for(i=1;i<=len;i++)
	{
		if(HT[i].weight<weight&&HT[i].parent==0)
		{
			s1=i;
			weight=HT[i].weight;
		}
	}
	weight=1000000;
	for(i=1;i<=len;i++)
	{
		if(HT[i].weight<weight&&HT[i].parent==0&&i!=s1)
		{
			s2=i;
			weight=HT[i].weight;
		}
	}
}
void CreateHuffmanTree(HuffmanTree &HT,int n)
{//创建哈夫曼树 
	int i,j,s1,s2;
	if(n<=1)
	    return ;
	int m=2*n-1;
	//HT=(HTNode)malloc((m+1)*sizeof(HTNode)); 
	HT=new HTNode[m+1];//new是c++中用于申请内存的运算符， 
	for(i=1;i<=m;i++)//将所有的节点初始化为0 
	{
		HT[i].parent=0;
		HT[i].lchild=0;
		HT[i].rchild=0;
	}
	printf("请输入你所要构造哈夫曼树的结点
");
	for(i=1;i<=n;i++)
	    scanf("%d",&HT[i].weight);//输入n个单元的叶子结点值 
    //创建树
	for(i=n+1;i<=m;i++)
	{
		Select(HT,i-1,s1,s2);
		HT[s1].parent=i;
		HT[s2].parent=i;//将s1,s2的双亲设置为i; 
		HT[i].lchild=s1;
		HT[i].rchild=s2;//将s1  s2设置为i的两个孩子 
		HT[i].weight=HT[s1].weight+HT[s2].weight;//生成新节点的值 
	} 
}
void CreatHuffmanCode(HuffmanTree HT,HuffmanCode &HC,int n)
{//求出哈夫曼编码 
	int i,start;
	int child,parent;
	HC=new char*[n+1];
	char cd[1100];
	for(i=1;i<=n;i++)
	{
		start=n-1;
		child=i;
		parent=HT[i].parent;
		while(parent!=0)
		{
			--start;
			if(HT[parent].lchild==child)
			cd[start]='0';
			else
			cd[start]='1';
			child=parent;
			parent=HT[parent].parent;
		}
		HC[i]=new char[n-start];
		strcpy(HC[i],&cd[start]);
	}
}
int query(HuffmanTree HT,int len,int edge)
{//查询你想要查找的哈夫曼编码的存储位置 
	int i;
	for(i=1;i<=len;i++)
		if(HT[i].weight==edge)
	return i;
}
int main()
{
	int k,n;
	HuffmanTree HT;
	HuffmanCode HC;
	printf("请输入您想要构造的哈夫曼树的结点个数
");
	scanf("%d",&n);
	CreateHuffmanTree(HT,n);
	printf("构造哈夫曼树完成
");
	CreatHuffmanCode(HT,HC,n);
	while(1)
	{
	    printf("请您输入一个你想要的查询哈夫曼编码的结点
");
	    scanf("%d",&k); 
	    printf("%s
",HC[query(HT,n,k)]);
	}

	return 0;
}