解决立即给分！！！！！

[zc_0702]

在一份电文中共使用5种字符：A，B，C，D，E，它们在电文中出现的频率分别为4，7，5，2，9。试通过构造最优化二叉树（要求左子树根结点的权值小于右子树根结点的权值）来确定每种字符的哈夫曼编码。

[zteliubin]

用程序计算吗，要花一点时间，结果应该是：
E
C B
D A

[ilmlb]

up

[Alicky]

UP

[xiaolongleee]

// huffman.cpp : 定义控制台应用程序的入口点。
//vs.net通过

#include "stdafx.h"
#define MAX 21

typedef struct
{
char data;
int weight;
int parent;
int left;
int right;
}Huffcode;

typedef struct
{
char cd[MAX];
int start;
}code;

int _tmain(int argc, _TCHAR* argv[])
{
Huffcode ht[2*MAX];
code hcd[MAX],d;
int i,k,f,l,r,n,c,m1,m2;

printf("元素个数：");
scanf("%d",&n);

for(i=1; i<=n; i++)
{
getchar();
printf("第%d个元素=>\n\t结点值:",i);
scanf("%c",&ht[i].data);
printf("\t权重：");
scanf("%d",&ht[i].weight);
}

for(i=1; i<=2*n-1; i++)
{
ht[i].parent = 0;
ht[i].left = 0;
ht[i].right = 0;
}

for(i=n+1; i<=2*n-1; i++)
{
m1 = m2 = 300;
l = r = 0;
for(k=1; k<=i-1; k++)
{
if(ht[k].parent == 0)
{
if(ht[k].weight < m1)
{
m2 = m1;
r = l;
m1 = ht[k].weight;
l = k;
}
else if(ht[k].weight < m2)
{
m2 = ht[k].weight;
r = k;
}
}
}

ht[l].parent = i;
ht[r].parent = i;
ht[i].weight = ht[l].weight + ht[r].weight;
ht[i].left = l;
ht[i].right = r;
}

for(i=1; i<=n; i++)
{
d.start = n+1;
c = i;
f = ht[i].parent;

while(f != 0)
{
if(ht[f].left == c)
{
d.cd[--d.start] = '0';
}
else
{
d.cd[--d.start] = '1';
}

c = f;
f = ht[f].parent;
}
hcd[i] = d;
}

printf("输出huffman编码：\n");

for(i=1; i<=n; i++)
{
printf("%c:",ht[i].data);

for(k=hcd[i].start; k<=n; k++)
{
printf("%c",hcd[i].cd[k]);
}
printf("\n");
}

return 0;
}

[soniczhm]

27
11 16
5(C) 6 7(B) 9(E)
2(D) 4(A)