数据结构：赫夫曼树和赫夫曼编码的存储表示

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程序：赫夫曼树和赫夫曼编码的存储表示

完成者：小单

完成时间：2013年5月26日

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//测试

#include <stdio.h>#include <stdlib.h>#include <string.h>#define  UINT_MAX 100  //假设各结点的权值不会超过100typedef struct{unsigned int weight; //权值unsigned int parent, lchild, rchild; }HTNode, *HuffmanTree;  //动态分配数组存储赫夫曼树typedef char **HuffmanCode;  //动态分配数组存储赫夫曼编码表int Min(HuffmanTree &HT,int i){//在HT[1...i]中选择parent为0且权值最小的结点//返回该结点的下标值//此函数被Select函数调用int j;unsigned int k = UINT_MAX;//假设各结点的权值不会超过UINT_MAXint flag;for(j = 1; j <= i; ++j){if(HT[j].weight < k && HT[j].parent == 0)//用父结点是否为0来判断此结点是否已经被选过{k = HT[j].weight;flag = j;}}HT[flag].parent = 1;//作个标记，说明已经被选择了，因为在Select函数中要选择权值小的两个结点return flag;}void Select(HuffmanTree &HT, int i, int &s1, int &s2){//在HT[1...i]中选择parent为0且权值最小的两个结点，其序号分别为s1,s2//s1 <= s2s1 = Min(HT,i);s2 = Min(HT,i);}void HuffmanCoding(HuffmanTree &HT, HuffmanCode &HC, int *w, int n){//w存放n个字符的权值（均>0），构造赫夫曼树HT，并求出n个字符的赫夫曼编码HCif(n <= 1)return ;int m = 2*n - 1; //严格的二叉树中，有n个叶子结点，就有2n - 1个结点HT = (HuffmanTree)malloc((m+1)*sizeof(HTNode));  //0号单元不用HuffmanTree p;int i;for(p = HT+1, i =1;i <= n; ++i, ++p, ++w){//初始化一棵赫夫曼树(*p).weight = *w;(*p).parent = 0;(*p).lchild = 0;(*p).rchild = 0;}for(;i<=m;++i,++p){(*p).weight = 0;(*p).parent = 0;(*p).lchild = 0;(*p).rchild = 0;}int s1,s2;//保存权值最小的两个结点的序号for(i=n+1; i <= m; ++i)//建立赫夫曼树{Select(HT,i-1,s1,s2);HT[s1].parent = i;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赫夫曼编码--------HC = (HuffmanCode)malloc((n+1)*sizeof(char**));  //分配n个字符编码的头指针向量,0号单元未用char *cd = (char *)malloc(n*sizeof(char));  //分配求编码的工作空间cd[n-1] = '\0';  //编码结束符int start;unsigned chld,father;for(i = 1; i <= n; ++i)  //逐个字符求赫夫曼编码{start = n-1;  //从编码结束符的位置开始for(chld=i, father=HT[i].parent; father!=0; chld=father, father=HT[father].parent)//从叶子到根逆向求编码{if(HT[father].lchild == chld)cd[--start] = '0';elsecd[--start] = '1';}HC[i] = (char*)malloc((n-start)*sizeof(char));  //为第i个字符编码分配空间strcpy(HC[i], &cd[start]);  //从cd复制编码到HC}free(cd);  //释放工作空间}int main(){HuffmanTree HT;HuffmanCode HC;printf("请输入权值的个数:\n");int n;scanf("%d",&n);int *w = (int*)malloc(n*sizeof(int));printf("请分别输入%d个权值:\n",n);int i;for(i=0; i<n; ++i){scanf("%d",w+i);}HuffmanCoding(HT, HC, w, n);printf("赫夫曼树为：\n");printf("NO\tweight\tparent\tlchild\trchild\n");for(i=1; i <= 2*n-1; ++i){printf("%d\t%d\t%d\t%d\t%d\n", i, HT[i].weight, HT[i].parent,HT[i].lchild, HT[i].rchild);}printf("赫夫曼编码为:\n");for(i = 1; i <= n; ++i)printf("%d|  |-->%s\n", i, HC[i]);return 0;}

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