05-树9 Huffman Codes（C）

日常，这一次，耗费我三天，其实第二天时便已经将 对整个框架有清晰的了解了，（看了解析了），但是一步步排除，确实让我学到了很多。

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 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

 面对这样的代码， 务必清晰

typedef struct TNode *HF;
typedef struct Heap *MinHeap;
struct TNode{
	int weight;
	HF Left;
	HF Right;
};
struct Heap{
	HF *data;
	int size;	
	int capacity;
};
MinHeap Init_Heap(int num)
{
	MinHeap H;
	H = (MinHeap)malloc(sizeof(struct Heap));
	H ->data = (HF*)malloc(sizeof(struct TNode) * (num + 1));
	H->data[0] = (HF)malloc(sizeof(struct TNode));
	H ->data[0] ->weight = -1;
	H ->data[0] ->Left = NULL;
	H ->data[0] ->Right = NULL;
	H ->size = 0;
	return H;
}

我，最开始对

H->data[0]=(Huffman)malloc(sizeof(struct TreeNode));

也不解，甚至写出了，

void attach(HF H, HF T)
{
	HF ->weight = T ->weight;
	HF ->Left = T ->Left;
	HF ->Right = T ->Right;
	free(T);
}
//这样的代码

后来明白了HF *data,可以视为 struct TNode **data;

H-> data 指分配了（num+1）大小的指针数组,

H ->data[0]指分配一个struct TNode空间,理解 struct TNode *data[0];（这里data[0]）

哈哈，C语言就是这样，指针我也只是粗略的了解，加油

 我的AC代码：

#include
#include
#include
#include
#include

typedef struct TNode *HF;
typedef struct Heap *MinHeap;
struct TNode{
	int weight;
	HF Left;
	HF Right;
};
struct Heap{
	HF *data;
	int size;	
	int capacity;
};

MinHeap Init_Heap(int num);
HF Init_Huffman();
void Insert_Heap(MinHeap H, HF T);
HF Pop_Node(MinHeap H);
HF Build_Huffman();
HF Build_Huffman(int num, int *code, char *ch);
int WPL(HF T, int depth);
bool Judge_Huffman(int num, int wpl, int *code, char *ch);

int main()
{
	int num;
	HF T;
	MinHeap H;
	char *ch;
	int *code, wpl;
	int test;
	scanf("%d", &num);
	ch = (char*)malloc(sizeof(char) * (num + 1));
	code = (int*)malloc(sizeof(int) * (num + 1));
	T = Build_Huffman(num, code, ch);
	wpl = WPL(T, 0);
	scanf("%d", &test);
	while(test--){
		if(Judge_Huffman(num, wpl, code, ch)){
			printf("Yes\n");
		}else{
			printf("No\n");
		}
	}
	return 0;
}
HF Build_Huffman(int num, int *code, char *ch)
{
	MinHeap H;
	HF T;
	int i, n;
	H = Init_Heap(num);
	for(i = 1; i <= num; i++){
		scanf(" %c %d", &ch[i], &code[i]);
		if(isalpha(ch[i]) && isupper(ch[i]) && code[i] > 0);
		else{
			printf("错误！请检查输入字符。\n");
		}
		T = Init_Huffman();
		T ->weight = code[i];
		Insert_Heap(H, T);
	}
	n = H->size;
	for(int i=1; i < n; i++)
	{
		T= Init_Huffman();
		T->Left = Pop_Node(H);
		T->Right = Pop_Node(H);
		T->weight = T->Left->weight+T->Right->weight;
		Insert_Heap(H,T);
	}
	T = Pop_Node(H);
	return T;
}
int WPL(HF T, int depth)
{
	if(!(T ->Left) && !(T ->Right)){
		return (T ->weight * depth);
	}else{
		return WPL(T ->Left, depth + 1) + WPL(T ->Right, depth + 1);
	}
}
MinHeap Init_Heap(int num)
{
	MinHeap H;
	H = (MinHeap)malloc(sizeof(struct Heap));
	H ->data = (HF*)malloc(sizeof(struct TNode) * (num + 1));
	H->data[0] = (HF)malloc(sizeof(struct TNode));
	H ->data[0] ->weight = -1;
	H ->data[0] ->Left = NULL;
	H ->data[0] ->Right = NULL;
	H ->size = 0;
	return H;
}
HF Init_Huffman()
{
	HF T;
	T = (HF)malloc(sizeof(struct TNode));
	T ->weight = 0;
	T ->Left = NULL;
	T ->Right = NULL;
	return T;
}
void Insert_Heap(MinHeap H, HF T)
{
	int i;
	i = ++(H->size);
	for(; T ->weight < H ->data[i/2]->weight; i /=2){
		H ->data[i] = H ->data[i/2];
	}
	H ->data[i] = T;
}
HF Pop_Node(MinHeap H)
{
	HF Temp, Min;
	int Child, Parent;
	Min = H ->data[1];
	Temp = H ->data[(H ->size)--];
	for(Parent = 1; Parent * 2 <= H ->size; Parent = Child){
		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 Min;
}
bool Judge_Huffman(int num, int wpl, int *code, char *ch)
{
	HF T, Temp;
	char c, *codes;
	int j, wgh;
	int length = 0;
	bool flag = true;
	T = Init_Huffman();
	codes = (char*)malloc(sizeof(char) * (num - 1));
	if(!codes) printf("没有分配成功\n");
	while(num --){
		scanf(" %c %s", &c, codes);
		for(j = 1; c != ch[j]; j++);
		wgh = code[j];
		Temp = T;
		for( j = 0; j < strlen(codes); j++){
			if(codes[j]=='0'){
				if(!Temp->Left)
					Temp->Left = Init_Huffman();
				Temp = Temp->Left;	
			}else if(codes[j] == '1'){
				if(!Temp->Right)
					Temp->Right = Init_Huffman();
				Temp = Temp->Right;
			}if(Temp->weight){
				flag = false;	
			}
		}
		if(Temp->Left || Temp->Right)
			flag = false;
		else{
			Temp->weight = wgh;
		}
		length += strlen(codes) * Temp->weight;
	}
	free(codes);
	if(length != wpl)
		flag = false;
	return flag;
}