poj 1521_huffmanb code

题目描述：

输入字符串，求用赫夫曼编码能得到的最好的字符编码方式，令总长度最小。求最小的长度。

解题思路：

建结构记录权重（字符出现的频率），父节点，左右子节点标识（n个叶子节点拥有不同的权重）作为赫夫曼树的叶子点结构；建立赫夫曼树(开一个2*n-1大小的数组，建立一棵有n个叶子的赫夫曼树，每次合并两个节点记录总权值)后，递归求得总权值，即为总长度。

时间复杂度为O(n2),n为不同的节点数。

代码：

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#define N 100000
 
typedef struct{
       int weight;
       int parent, lc, rc;
}HTNode;

HTNode ht[N];
char str[N];
int sum1,sum, n, num[27], i,j, min, tmp1,tmp2;
double sum2;

void sum_weight(HTNode hn, int i)
{
   if(hn.lc == 0&& hn.rc == 0)
     sum += hn.weight*i;
   else
   {
      sum_weight(ht[hn.lc], i+1);
      sum_weight(ht[hn.rc], i+1);
   }
}

main()
{
   scanf("%s", str);
   while(strcmp(str, "END") !=0)
   {
     n = strlen(str);
     sum1 = n*8;
     for(i=0;i<=26;i++)
        num[i] = 0;
     for(i=0;i<n;i++)
     {
        if(str[i] == '_')
           num[26]++;
        else
           num[str[i]-'A']++;
     }
     j=0;
     for(i=0;i<27;i++)
     {
        if(num[i]!=0)
        {
           ht[j].weight = num[i];
           ht[j].parent = 0;
           ht[j].lc = 0;
           ht[j].rc = 0;
           j++;
        }
     }
     n = j; 
     //create hurff tree
     for(i=n; i<2*n-1; i++)
     {
        //select less two nodes
        min = 10000;
        for(j=0;j<i;j++)
        {
           if(ht[j].parent == 0 &&ht[j].weight < min)
           {
              tmp1 = j;
              min = ht[j].weight;
           }
        }
        min = 10000;
        for(j=0;j<i;j++)
        {
           if(ht[j].parent == 0 &&ht[j].weight < min&& j!= tmp1)
           {
              tmp2 = j;
              min = ht[j].weight;
           }
        }
        //更改节点信息:
        ht[tmp1].parent = i;
        ht[tmp2].parent = i;
        ht[i].lc = tmp1;
        ht[i].rc = tmp2;
        ht[i].weight = ht[tmp1].weight + ht[tmp2].weight;
        ht[i].parent = 0;
     }
     sum = 0;
     //calculate the sum of weight, root  is in2*n-2
     if(n == 1)
        sum = strlen(str);
     else
        sum_weight(ht[2*n-2],0);
     sum2 = (double)((double)sum1/(double)sum);
     printf("%d %d %.1f\n",sum1, sum,sum2);
     
     scanf("%s", str);
   }
   //system("pause");
   return 0;
}