05-树9 Huffman Codes

05-树9 Huffman Codes (30分)

In 1953, David A. Huffman published his paper "A Method for the Construction of Minimum-Redundancy Codes", and hence printed his name in the history of computer science. As a professor who gives the final exam problem
on Huffman codes, I am encountering a big problem: the Huffman codes are NOT unique. For example, given a string "aaaxuaxz", we can observe that the frequencies of the characters 'a', 'x', 'u' and 'z' are 4, 2, 1 and 1, respectively. We may either encode the
symbols as {'a'=0, 'x'=10, 'u'=110, 'z'=111}, or in another way as {'a'=1, 'x'=01, 'u'=001, 'z'=000}, both compress the string into 14 bits. Another set of code can be given as {'a'=0, 'x'=11, 'u'=100, 'z'=101}, but {'a'=0, 'x'=01, 'u'=011, 'z'=001} is NOT
correct since "aaaxuaxz" and "aazuaxax" can both be decoded from the code 00001011001001. The students are submitting all kinds of codes, and I need a computer program to help me determine which ones are correct and which ones are not.

Input Specification:

Each input file contains one test case. For each case, the first line gives an integer NN (2\le
N\le 632≤N≤63),
then followed by a line that contains all the NNdistinct
characters and their frequencies in the following format:

c[1] f[1] c[2] f[2] ... c
f

where

c[i]

is
a character chosen from {'0' - '9', 'a' - 'z', 'A' - 'Z', '_'}, and

f[i]

is
the frequency of

c[i]

and is
an integer no more than 1000. The next line gives a positive integer MM (\le
1000≤1000),
then followed by MMstudent
submissions. Each student submission consists of NN lines,
each in the format:

c[i] code[i]

where

c[i]

is
the

i

-th character and

code[i]

is
an non-empty string of no more than 63 '0's and '1's.

Output Specification:

For each test case, print in each line either "Yes" if the student's submission is correct, or "No" if not.
Note: The optimal solution is not necessarily generated by Huffman algorithm. Any prefix code with code length being optimal is considered correct.

Sample Input:

7
A 1 B 1 C 1 D 3 E 3 F 6 G 6
4
A 00000
B 00001
C 0001
D 001
E 01
F 10
G 11
A 01010
B 01011
C 0100
D 011
E 10
F 11
G 00
A 000
B 001
C 010
D 011
E 100
F 101
G 110
A 00000
B 00001
C 0001
D 001
E 00
F 10
G 11

Sample Output:

Yes
Yes
No
No

题目思路：

就是何老师讲的霍夫曼树，关键是前缀码的判断，这点参考了inaho的思路

#include<stdio.h>
#include <stdlib.h>

#define MinData 0

typedef struct TreeNode *HuffmanTree;
struct TreeNode{
int weight;
HuffmanTree left;
HuffmanTree right;
};
//霍夫曼树

typedef struct HeapStruct *MinHeap;
struct HeapStruct{
HuffmanTree elements;
int size;
int capacity;
};
//最小堆

MinHeap BuildMinHeap(int Maxsize,int weight[]);
MinHeap CreateMinHeap (int Maxsize);
void Insert(MinHeap H,HuffmanTree HT);
HuffmanTree DeleteMin(MinHeap H);
HuffmanTree Huffman(MinHeap H);
void GetWpl(HuffmanTree HT,int layer,int *wpl);

int code_length(char *a);
int compare(char *c1, char *c2);

int main(int argc, char const *argv[])
{
//freopen("test.txt","r",stdin);

int N;
scanf("%d\n",&N);//一共几个编码

char c
;//记录字符
int f
;//记录字符对应权重

for(int i = 0; i < N; i++)
{
if(i == N - 1)
{
scanf("%c %d", &c[i], &f[i]);
}
else{
scanf("%c %d ", &c[i], &f[i]);
}

}

MinHeap MH = BuildMinHeap(N,f);
HuffmanTree HT = Huffman(MH);

int minwpl = 0;
GetWpl(HT,0,&minwpl);

//读入有多少个编码组合
int M;
scanf("%d\n",&M);
char ch
,code
[64];

for(int j = 0; j < M; j++)//一共M组
{
for(int i = 0; i < N; i++)///每组N个数
{
scanf("%c %s\n",&ch[i],&code[i]);
}

int flag = 1;//1表示是前缀码//前缀码判断

for (int i = 0; i < N; i++)
{
for (int k = i+1; k < N; ++k)
{
if(compare(code[i], code[k]))
{
if(flag)
printf("No\n");
flag = 0;
}
}
}

int stu_wpl = 0;  //比较wpl
if( flag )
{
for (int i = 0; i < N; ++i)
stu_wpl += f[i]*code_length(code[i]);
if(minwpl == stu_wpl)
printf("Yes\n");
else
printf("No\n");
}

}
return 0;

}

int code_length(char *a)
{
int len = 0;
char *p = a;
while(*p != '\0'){
p++;
len++;
}
return len;
}

int compare(char *c1, char *c2)
{
char *a = c1, *b = c2;
while(*a!='\0' && *b!='\0'){
if(*a != *b)
return 0;
a++;
b++;
}
return 1;
}

MinHeap BuildMinHeap(int Maxsize,int weight[])
{
int i;
MinHeap H = CreateMinHeap(Maxsize);
HuffmanTree temp = (HuffmanTree)malloc(sizeof(struct TreeNode));
for(int i = 0; i < Maxsize; i++)
{
temp->weight = weight[i];
temp->left = temp->right = NULL;
Insert(H,temp);
}
free(temp);
return H;
}

MinHeap CreateMinHeap (int Maxsize)
{
MinHeap H = (MinHeap)malloc(sizeof(struct HeapStruct));
H->elements = (HuffmanTree)malloc(sizeof(struct TreeNode)*(Maxsize+1));
//因为Elemens[0]作为哨兵，从[1]开始存放，所以分配MaxSize+1空间
H->size = 0;
H->capacity = Maxsize;
H->elements[0].weight= MinData;//哨兵
return H;
}

void Insert(MinHeap H,HuffmanTree HT)
{
int i;
//将X插入H
if(H->size == H->capacity){
printf("最大堆已满！");
return;
}
i = ++(H->size); //i指向插入后堆中的最后一个元素的位置(该结点此时为空结点)
for( ; H->elements[i/2].weight > HT->weight ; i/=2)
//堆从1开始
//不断和父节点比较，父节点大，就往下走
{
H->elements[i].weight = H->elements[i/2].weight;
}
H->elements[i] = *HT;//将X插入；
}

HuffmanTree DeleteMin(MinHeap H)
{
//从最小堆H中取出键值为最小的元素，并删除一个结点
int parent, child;
HuffmanTree MinItem, temp;

MinItem = (HuffmanTree)malloc(sizeof(struct TreeNode));
temp = (HuffmanTree)malloc(sizeof(struct TreeNode));

if ( H->size == 0)
{
printf("最小堆已为空");
return NULL;
}

*MinItem = H->elements[1];//保存最小的元素
*temp = H->elements[H->size--];//从最后一个元素插到顶点来比较

for(parent=1;parent*2 <= H->size; parent = child)//有没有左儿子
{
child = parent * 2;//有的话比较左儿子
if ((child != H->size)&&
(H->elements[child].weight > H->elements[child+1].weight))
//比较左右儿子哪个小
{
child ++;
}
if(temp->weight <= H->elements[child].weight)
{
break;
}
else
{
H->elements[parent] = H->elements[child];
}
}
H->elements[parent] = *temp;
free(temp);
return MinItem;
}

HuffmanTree Huffman(MinHeap H)
{
int i; HuffmanTree HT;
int times = H->size;//H->size的值会发生变化，所以要用另一个变量来存储
for( i = 1; i < times ; i++)
{
HT = (HuffmanTree)malloc(sizeof(struct TreeNode));
HT->left = DeleteMin(H);
HT->right = DeleteMin(H);
HT->weight = HT->left->weight + HT->right->weight;
Insert(H,HT);
}
HT = DeleteMin(H);
return 	HT;
}

void GetWpl(HuffmanTree HT,int layer,int *wpl)
{
if(HT->left== NULL && HT->right == NULL)
{
(*wpl) = (*wpl)  + layer * HT->weight;
}
else//不是叶节点
{
GetWpl(HT->left,layer+1,wpl);
GetWpl(HT->right,layer+1,wpl);
}
}