【Data Structures and Algorithms】7-9 Huffman Codes（30 分）

 

7-9 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≤N≤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 (≤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

 

思路主要有几点：

1、每个字母的编码长度都<=n-1, n是字母个数，也就是题目中的N。

2、Huffman树不唯一，但是带权路径长度（WPL）是唯一的，根据第二行给出的频率计算WPL与之后每组输入的WPL对比。           （另一个方法：WPL=所有非叶节点的权值之和）

3、建树判断是否有前缀码。

4、判断是否有不同字母的编码相同，这种情况下去判断前缀码是无效的，因为它们在树中的位置是重合的，

      这一点是我一开始没有考虑到的一点。

需要注意的有：

1、输入的字符顺序并不是乱的，是按给出频率的顺序输入的。

2、scanf读入数据时，格式很重要，不要漏了\n，\n的位置也很重要。

 

#include
#include
#include

#define maxsize 65
typedef struct treenode* Tree;
typedef Tree intArray[maxsize];
typedef struct HNode* MinHeap;
typedef char str[maxsize];
struct HNode {
	intArray Data;
	int size;
};
struct treenode {
	int freq; 
	Tree left;
	Tree right;
};
struct string {
	char str[maxsize];
	int length;
};

MinHeap CreateHeap(int n,int F[]) {
	MinHeap H=(MinHeap)malloc( sizeof(struct HNode) );
	int i;
	char b;
	H->size=0;
	for(i=0;i<=n;i++) {
		Tree T=(Tree)malloc( sizeof(struct treenode) );
		T->freq=0;
		T->left=T->right=NULL;
		H->Data[i]=T;
	}
	scanf("%c %d",&b,&F[1]); 
	H->Data[++H->size]->freq=F[1];
	for(i=2;i<=n;i++) {
		scanf(" %c %d",&b,&F[i]);
		H->Data[++H->size]->freq=F[i];
	}
	
	return H;
}

void PercDown(MinHeap H,int parent) {
	int child;
	Tree x;
	x=H->Data[parent];
	for(;parent*2<=H->size;parent=child) {
		child=parent*2;
		if(child+1<=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;
	
}
void BuildMinHeap(MinHeap H) {
	int i;
	for(i=H->size/2;i>0;i--) {
		PercDown(H,i);
	}
}
Tree Delete(MinHeap H) {
	Tree min=NULL;
	if(H->size) {
		min=H->Data[1];
		H->Data[1]=H->Data[H->size--];
	}
	if(H->size) 
		PercDown(H,1);
		
	return min;
}
void Insert(MinHeap H,Tree T) {
	int parent,child;
	for(child=++H->size,parent=child/2;parent>0;child=parent,parent/=2) {
		if(H->Data[parent]->freq>T->freq )
			H->Data[child]=H->Data[parent];
		else
			break;
	}
	H->Data[child]=T;	
}
Tree Huffman(MinHeap H) {
	int i,n=H->size;
	Tree T=NULL;
	for(i=1;ileft=Delete(H);
		T->right=Delete(H);
		T->freq=T->left->freq + T->right->freq;
		Insert(H,T);
	}
	return T;
}
int WPL(Tree T,int depth) {
	if(T->left!=NULL)
		return WPL(T->left,depth+1)+WPL(T->right,depth+1);
	else
		return T->freq*depth;
}
Tree BuildTree(int num,str* Q[],int F[]) {
	int j,m;
	Tree root=NULL,parent=NULL;
	Tree prefixTree=(Tree)malloc( sizeof(struct treenode) );
	prefixTree->freq=0;
	prefixTree->left=prefixTree->right=NULL;
	root=prefixTree;
	for(j=1;j<=num;j++) {
		parent=root;
		for(m=0;(*Q[j])[m]!='\0';m++) {
			if( (*Q[j])[m] == '0') {
				if(!parent->left) {
					prefixTree=(Tree)malloc( sizeof(struct treenode) );
					prefixTree->freq=0;
					prefixTree->left=prefixTree->right=NULL;
					parent->left=prefixTree;
				}
				parent=parent->left;

			}
			else if( (*Q[j])[m] == '1') {
				if(!parent->right) {
					prefixTree=(Tree)malloc( sizeof(struct treenode) );
					prefixTree->freq=0;
					prefixTree->left=prefixTree->right=NULL;
					parent->right=prefixTree;
				}
				parent=parent->right;

			}

		}
		parent->freq=F[j];

	} 

	return root;
}
void Destroy(Tree T) {
	if(T) {
		Destroy(T->left);
		Destroy(T->right);
		free(T);
	}

}
int TreeTravel(Tree root,int flag) {
	if(root && !flag) {
		flag=TreeTravel(root->left,flag);
		flag=TreeTravel(root->right,flag);
		if( root->freq!=0 && (root->left!=NULL || root->right!=NULL) )
			flag=1;
			
	}
	return flag;
}

int Stringcompare(str* Q[],int n) {
	int i,j,flag;
	for(i=1;inum-1) 
				flag=1;
			else 
				sum+=strlen(*Q[j])*F[j];
		}
		
		if(flag) printf("No\n");
		else if(!Stringcompare(Q,num)) printf("No\n");
		else if(sum!=wpl) {
			printf("No\n");
		}
		else {
		 	root=BuildTree(num,Q,F);
			flag=TreeTravel(root,0);
			if(flag) printf("No\n");
			else printf("Yes\n");
			Destroy(root);
		}  
			
	}

	//system("pause");
	return 0;
}