简单哈弗曼编码

//此代码是数据结构的原始模板，可以刚接触或考研时借鉴下，不适于刷题
#include<stdio.h>
#include<string.h>
#include<stdlib.h>
int s1,s2;
typedef struct node
{
	int weight;
	int parent;
	int lchild;
	int rchild;
}tree,*bitree;
void selet(tree ht[],int x)
{
	int i,min1='0x3f',min2='0x3f';
	for(i=1;i<=x;i++)
	{
       if(min1>ht[i].weight&&ht[i].parent==0)
	   {
	     min1=ht[i].weight;
	   }
	}
   for(i=1;i<=x;i++)
	{
       if(min1==ht[i].weight&&ht[i].parent==0)
	   {
	     s1=i;
		 ht[i].parent=-1;
		 break;
	   }
	}
	for(i=1;i<=x;i++)
	{
       if(min2>ht[i].weight&&ht[i].parent==0)
	   {
	     min2=ht[i].weight;
	   }
	}
		for(i=1;i<=x;i++)
	{
       if(min2==ht[i].weight&&ht[i].parent==0)
	   {
	     s2=i;
		 break;
	   }
	}
	ht[s1].parent=0;
	if(min1==min2)
	{
		int r=s1;
		s1=s2;
		s2=r;
	}
	
}
bitree crthuffmantree(tree ht[],int w[],int n)//构造哈弗曼树
{
	for(int i=1;i<=n;i++)
	{
		ht[i].weight=w[i];
		ht[i].parent=0;
		ht[i].lchild=0;
		ht[i].rchild=0;
	}
	int m=2*n-1;
	for(i=n+1;i<=m;i++)
	{
		ht[i].weight=0;
	    ht[i].parent=0;
		ht[i].lchild=0;
		ht[i].rchild=0;
	}
	for(i=n+1;i<=m;i++)
	{
		selet(ht,i-1);
	    ht[i].weight=ht[s1].weight+ht[s2].weight;
	    ht[s1].parent=i;
		ht[s2].parent=i;
		ht[i].lchild=s1;
		ht[i].rchild=s2;
	}
	return ht;
}
typedef char *huffmancode[300];
void crthuffmancode(tree ht[],huffmancode hc,int n,char is[])
{
	char *cd;
	int start,c,p,i;
	cd=(char*)malloc(n*sizeof(char));
	cd[n-1]='\0';
	for(i=1;i<=n;i++)
	{
		start=n-1;
		c=i;
		p=ht[i].parent;
		if(p==0)
		{
		hc[i]=(char*)malloc((n-start)*sizeof(char));
		strcpy(hc[i],"1");
		}
		else
		{
		while(p!=0)
		{
			--start;
			if(ht[p].lchild==c)
				cd[start]='0';
			else
				cd[start]='1';
			c=p;
			p=ht[p].parent;
		}
		hc[i]=(char*)malloc((n-start)*sizeof(char));
		strcpy(hc[i],&cd[start]);
		}
	}
	for(i=1;i<=n;i++)
	printf("%c:%s\n",is[i],hc[i]);
	free(cd);
}
int main()
{
	char english[10000],is[100];
	huffmancode Huffmancode;
	printf("请输入一字符串可利用哈弗曼树原理求出字符串中各字符的哈弗曼编码:\n");
	scanf("%s",english);
	int len=strlen(english);
	int i,j,ajj[300];
	ajj[0]=0;
	ajj[1]=1;
	is[1]=english[0];
	int ans=1;
	for(i=1;i<len;i++)
	{
		for(j=1;j<=ans;j++)
		{
		if(english[i]==is[j])
		{
			ajj[j]++;
			break;
		}
		}
		if(j==ans+1)
		{
			is[ans+1]=english[i];
			ajj[ans+1]=1;
			ans++;//表示有多少个不同的字母
		}
	}
	tree hufu[400];
	bitree ht1;
	hufu[0].weight=0;
	hufu[0].parent=0;
	hufu[0].lchild=0;
	hufu[0].rchild=0;
	ht1=crthuffmantree(hufu,ajj,ans);
	crthuffmancode(ht1,Huffmancode,ans,is);
	return 0;
}