05-树9 Huffman Codes

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 N (2≤N≤63),
then followed by a line that contains all the Ndistinct
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 M (≤1000),
then followed by Mstudent
submissions. Each student submission consists of N 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

123

思路：

题目意思是给定一组字母和出现的频率，再给出几组编码，判断是否符合哈夫曼编码规则

我一开始想是否能不通过构造huffman树来做，但是想不出一个可以不建树就能获得带权路径长度（WPL）的方法

#include <iostream>
#include <cstdio>
#include <vector>
#include <string>
using namespace std;
#define MAXSIZE 64
int nodenum;
int c[64];
char f[64];
typedef struct TNode* HuffTree;
struct TNode
{
HuffTree left;
HuffTree right;
int freq;
};

struct heap
{
HuffTree* data;
int size;
int capacity;
};
typedef struct heap* Minheap;

Minheap CreatHeap()
{
Minheap H = new struct heap;
H->data = new HuffTree[MAXSIZE];
H->size = 0;
H->capacity = MAXSIZE;
H->data[0] = new struct TNode;
H->data[0]->freq = -1;
return H;
}
bool Insert( Minheap H, HuffTree f )
{
int i;
if ( H->size == H->capacity ){
printf("minheap is full");
return false;
}
i = ++H->size;
for ( ; H->data[i/2]->freq > f->freq; i/=2 )
H->data[i] = H->data[i/2];
H->data[i] = f;
return true;
}
bool IsEmpty( Minheap H )
{
return (H->size == 0);
}

HuffTree DeleteMin( Minheap H )
{
int Parent, Child;
HuffTree MinItem, X;

if ( IsEmpty(H) ) {
printf("minheap is empty");
}

MinItem = H->data[1];
X = H->data[H->size--];
for( Parent=1; Parent*2<=H->size; Parent=Child ) {
Child = Parent * 2;
if( (Child!=H->size) && (H->data[Child]->freq > H->data[Child+1]->freq) )
Child++;
if( X->freq <= H->data[Child]->freq ) break;
else
H->data[Parent] = H->data[Child];
}
H->data[Parent] = X;

return MinItem;
}
HuffTree HuffmanTree()
{
Minheap h = CreatHeap();
for(int i=0; i<nodenum; i++)
{

HuffTree f = new struct TNode;
f->freq = c[i];
f->left = NULL;
f->right = NULL;
if(!Insert(h,f))
break;
}
for(;;)
{
if(h->size == 1) break;
HuffTree f = new struct TNode;
f->left = DeleteMin(h);
f->right = DeleteMin(h);
f->freq = f->left->freq + f->right->freq;
Insert(h,f);//printf("fleft=%d,fright=%d,ffreq=%d\n",f->left->freq,f->right->freq,f->freq);
}
return DeleteMin(h);
}
int WPL(HuffTree tree,int depth)
{
if((!tree->left)&&(!tree->right))
{
//printf("depth:%d,leaf->freq:%d\n",depth,tree->freq);
return depth*(tree->freq);
}
else
{
//printf("depth:%d,tree->freq:%d\n",depth,tree->freq);
return WPL(tree->left,depth+1)+WPL(tree->right,depth+1);
}
}

bool check(HuffTree tree,string s)
{
bool flag = false;
HuffTree p = tree;
for(int i=0;i<s.size();i++)
{
if(s[i]=='0')
{
if(!p->left)
{
p->left = new TNode;
p->left->left = NULL;
p->left->right = NULL;
p = p->left;
flag=true;
}
else
{
p = p->left;
}
}
else if(s[i]=='1')
{
if(!p->right)
{
p->right = new TNode;
p->right->left = NULL;
p->right->right = NULL;
p = p->right;
flag=true;
}
else
{
p = p->right;
}
}
}
return flag;
}

int main()
{
int case_;
cin>>nodenum;
for(int i=0;i<nodenum;i++)
{
cin>>f[i]>>c[i];
}
HuffTree tree = HuffmanTree();
int wpl = WPL(tree, 0);
cin>>case_;
for(int j=0;j<case_;j++)
{
HuffTree root = new TNode;
root->left = NULL;
root->right = NULL;
int s_wpl=0;
string judge="";
for(int i=0;i<nodenum;i++)
{
char ch;
string s;
cin>>ch>>s;
if(s.size()>nodenum-1){
judge="No";
break;
}
s_wpl += s.size()*c[i];
if(!check(root,s)) judge="No";
}
if(judge.empty()&&s_wpl==wpl)judge = "Yes";
else judge="No";
cout<<judge<<endl;
}

return 0;
}