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 $N$ ($2\le N\le 63$), then followed by a line that contains all the $N$distinct characters and their frequencies in the following format:

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

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 $M$ ($\le 1000$), then followed by $M$ student 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

主要思路：

1、通过小根堆构建Huffman树

2、通过Huffman树计算最小编码长度

3、判断输入的编码是否符合要求（1、编码长度与Huffman编码长度相同 2、前缀编码）

#include<iostream>
#include <string>
using namespace std;
#define MaxNum 64struct TreeNode//树的结点
{int Weight=0;TreeNode *Left=nullptr;TreeNode *Right=nullptr;
};struct HeapNode//堆
{TreeNode Data[MaxNum];int Size=0;
};HeapNode *CreateHeap(int N)//创建一个新的小根堆
{HeapNode *H=new(HeapNode);H->Data[0].Weight=-1;return H;
}TreeNode *DeleteMin(HeapNode *H)//从堆中删除一个结点
{int Parent=0,Child=0;TreeNode temp;TreeNode *MinItem=new(TreeNode);*MinItem=H->Data[1];temp=(H->Data[(H->Size)--]);for (Parent = 1; Parent*2 <= H->Size ; Parent=Child)//寻找删除结点前堆中最后一个结点在新堆中的插入位置{Child=Parent*2;if ((Child!=H->Size)&&((H->Data[Child].Weight)>(H->Data[Child+1].Weight))){Child++;}if ((temp.Weight)<=(H->Data[Child].Weight)){break;}else{H->Data[Parent]=H->Data[Child];}}H->Data[Parent]=temp;return MinItem;}void Insert(HeapNode *H,TreeNode *item)//插入新结点到堆中
{int i=0;i=++(H->Size);for (;H->Data[i/2].Weight>item->Weight; i/=2){H->Data[i]=H->Data[i/2];}H->Data[i]=*item;}HeapNode *ReadData(int N,HeapNode *H,int A[])//读取各个节点的权值输入数据
{char s='\0';int value=0;for (int i=0; i<N; ++i){cin>>s;cin>>value;A[i]=value;TreeNode *T=new(TreeNode);T->Weight=value;Insert(H, T);}return H;
}TreeNode *Huffman(HeapNode *H)//构建Huffman树
{TreeNode *T=nullptr;int num=H->Size;for (int i=0; i<num-1; ++i){T=new(TreeNode);T->Left=DeleteMin(H);T->Right=DeleteMin(H);T->Weight=T->Left->Weight+T->Right->Weight;Insert(H, T);}T=DeleteMin(H);return T;
}int WPL(TreeNode *T,int Depth)//计算Huffman树的编码长度
{if ((T->Left==nullptr)&&(T->Right==nullptr)){return Depth*(T->Weight);}else{return (WPL(T->Left,Depth+1)+WPL(T->Right,Depth+1));}
}
struct JNode
{int Flag=0;JNode *Left=nullptr;JNode *Right=nullptr;};
bool Judge(string S,JNode *J)//判断该次编码能否符合前缀编码的要求
{int i=0;for (; i<S.length(); ++i){if (S[i]=='0'){if (J->Left==nullptr){JNode *J_1=new(JNode);J->Left=J_1;}else{if (J->Left->Flag==1){return false;}}J=J->Left;}else{if (J->Right==nullptr){JNode *J_1=new(JNode);J->Right=J_1;}else{if (J->Right->Flag==1){return false;}}J=J->Right;}}J->Flag=1;if (J->Left==nullptr&&J->Right==nullptr){return true;}else{return false;}
}int main(int argc, char const *argv[])
{int N=0,n=0;cin>>N;HeapNode *H=CreateHeap(N);int Value[MaxNum]={};H=ReadData(N,H,Value);TreeNode *T=Huffman(H);int CodeLen=WPL(T,0);cin>>n;string temp="\0";char c='\0';bool result=false;for (int i=0; i<n; ++i){int count=0,flag=0;JNode *J=new(JNode);for (int k=0; k<N; ++k){cin>>c>>temp;count+=temp.length()*Value[k];if (!flag){result=Judge(temp,J);if (!result){flag=1;}}}delete J;if (result&&(count==CodeLen))//前缀编码且编码长度之和与Huffman编码相同{cout<<"Yes"<<endl;}else{cout<<"No"<<endl;}}return 0;
}