哈夫曼树的创建 实验报告

  1 #include<string.h>
  2 #include<stdlib.h>
  3 #include<stdio.h>
  4 
  5 int m,s1,s2;
  6 
  7 typedef struct
  8 {
  9     unsigned int weight;
 10     unsigned int parent,lchild,rchild;
 11 } HTNode,*HuffmanTree; //动态分配数组存储哈夫曼树
 12 typedef char *HuffmanCode; //动态分配数组存储哈夫曼编码表
 13 
 14 void Select(HuffmanTree HT,int n)
 15 {
 16     int i,j;
 17     for(i = 1; i <= n; i++)
 18         if(!HT[i].parent)
 19         {
 20             s1 = i;
 21             break;
 22         }
 23     for(j = i+1; j <= n; j++)
 24         if(!HT[j].parent)
 25         {
 26             s2 = j;
 27             break;
 28         }
 29     for(i = 1; i <= n; i++)
 30         if((HT[s1].weight>HT[i].weight)&&(!HT[i].parent)&&(s2!=i))
 31             s1=i;
 32     for(j = 1; j <= n; j++)
 33         if((HT[s2].weight>HT[j].weight)&&(!HT[j].parent)&&(s1!=j))
 34             s2=j;
 35 }
 36 
 37 void HuffmanCoding(HuffmanTree &HT, HuffmanCode HC[], int *w, int n)
 38 {
 39 // 算法6.13
 40 // w存放n个字符的权值(均>0)，构造哈夫曼树HT，
 41 // 并求出n个字符的哈夫曼编码HC
 42     int i, j;
 43     char *cd;
 44     int p;
 45     int cdlen;
 46 
 47     if (n<=1)
 48         return;
 49     m = 2 * n - 1;
 50     HT = (HuffmanTree)malloc((m+1) * sizeof(HTNode)); // 0号单元未用
 51     for (i=1; i<=n; i++)   //初始化
 52     {
 53         HT[i].weight=w[i-1];
 54         HT[i].parent=0;
 55         HT[i].lchild=0;
 56         HT[i].rchild=0;
 57     }
 58     for (i=n+1; i<=m; i++)   //初始化
 59     {
 60         HT[i].weight=0;
 61         HT[i].parent=0;
 62         HT[i].lchild=0;
 63         HT[i].rchild=0;
 64     }
 65     puts("\n哈夫曼树的构造过程如下所示：");
 66     printf("HT初态:\n 结点 weight parent lchild rchild");
 67     for (i=1; i<=m; i++)
 68         printf("\n%4d%8d%8d%8d%8d",i,HT[i].weight,
 69                HT[i].parent,HT[i].lchild, HT[i].rchild);
 70     for (i=n+1; i<=m; i++)   // 建哈夫曼树
 71     {
 72 // 在HT[1..i-1]中选择parent为0且weight最小的两个结点，
 73 // 其序号分别为s1和s2。
 74         Select(HT, i-1);
 75         HT[s1].parent = i;
 76         HT[s2].parent = i;
 77         HT[i].lchild = s1;
 78         HT[i].rchild = s2;
 79         HT[i].weight = HT[s1].weight + HT[s2].weight;
 80         printf("\nselect: s1=%d s2=%d\n", s1, s2);
 81         printf(" 结点 weight parent lchild rchild");
 82         for (j=1; j<=i; j++)
 83             printf("\n%4d%8d%8d%8d%8d",j,HT[j].weight,
 84                    HT[j].parent,HT[j].lchild, HT[j].rchild);
 85 
 86     }
 87 
 88 //------无栈非递归遍历哈夫曼树，求哈夫曼编码
 89     cd = (char *)malloc(n*sizeof(char)); // 分配求编码的工作空间
 90     p = m;
 91     cdlen = 0;
 92     for (i=1; i<=m; ++i) // 遍历哈夫曼树时用作结点状态标志
 93         HT[i].weight = 0;
 94     while (p)
 95     {
 96         if (HT[p].weight==0)   // 向左
 97         {
 98             HT[p].weight = 1;
 99             if (HT[p].lchild != 0)
100             {
101                 p = HT[p].lchild;
102                 cd[cdlen++] =‘0‘;
103             }
104             else if (HT[p].rchild == 0)   // 登记叶子结点的字符的编码
105             {
106                 HC[p] = (char *)malloc((cdlen+1) * sizeof(char));
107                 cd[cdlen] =‘\0‘;
108                 strcpy(HC[p], cd); // 复制编码(串)
109             }
110         }
111         else if (HT[p].weight==1)     // 向右
112         {
113             HT[p].weight = 2;
114             if (HT[p].rchild != 0)
115             {
116                 p = HT[p].rchild;
117                 cd[cdlen++] =‘1‘;
118             }
119         }
120         else     // HT[p].weight==2，退回退到父结点，编码长度减1
121         {
122             HT[p].weight = 0;
123             p = HT[p].parent;
124             --cdlen;
125         }
126     }
127 } // HuffmanCoding
128 int main()
129 {
130     HuffmanTree HT;
131     HuffmanCode *HC;
132     int *w,n,i;
133     puts("输入结点数:");
134     scanf("%d",&n);
135     HC = (HuffmanCode *)malloc(n*sizeof(HuffmanCode));
136     w = (int *)malloc(n*sizeof(int));
137     printf("输入%d个结点的权值\n",n);
138     for(i = 0; i < n; i++)
139         scanf("%d",&w[i]);
140     HuffmanCoding(HT,HC,w,n);
141     puts("\n各结点的哈夫曼编码:");
142     for(i = 1; i <= n; i++)
143         printf("%2d(%4d):%s\n",i,w[i-1],HC[i]);
144     getchar();
145 }