05-树9 Huffman Codes (30 分)

题目

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

题目理解

就是检查学生输入的编码是否合理，一是WPL合理，二是前缀码，不是哈夫曼编码也可以。

思路

由两方面进行判断：WPL长度和是否是前缀码。
一开始没弄清前缀码是啥就直接对提交编码进行字符串方面的比较…后面提交只有一个测试点通过才去查了定义，草。

检查编码是否是前缀码的方法是根据编码建一棵二叉树，建二叉树的时候会遍历各个节点，当遍历经过了某个节点是另一段编码的终点，就说明那段编码是这段正在遍历的编码的前缀（难以表述，意会就好）

代码

#include
#include
#include
#pragma warning(disable:4996)
#define MAXN 65
typedef struct HaffCode *PtrH;
typedef struct Min_Heap *PtrM;
typedef struct Submit *PtrS;
typedef struct OrigData *PtrO;
typedef struct JNode *PtrJ;
struct HaffCode
{
	char rep;
	int wei;
	PtrH right, left;
};
struct Min_Heap
{
	int size;
	int cap;
	PtrH Elems[MAXN];
};
struct Submit
{
	char rep;
	char code[MAXN];
	int length;
};
struct OrigData
{
	char rep;
	int weight;
};
struct JNode 
{
	int flag;
	PtrJ left;
	PtrJ right;
};
void Insert(PtrM M, PtrH item);
PtrH Delete(PtrM M);
PtrM Create(int maxsize);
void Establish(PtrM M,PtrO Orig);
void PercDown(PtrM M, int index);
PtrH Huffman(PtrM M);
int WPL(PtrH H, int height);
void Print(PtrH H);
void Check(int WPL, int N, PtrO Orig);
int FindWeight(PtrO Orig, char rep,int N);
PtrJ NewNode(void);
int Judge(char *s, PtrJ J);


int main()
{
	PtrM M;
	PtrH H;
	OrigData Orig[MAXN];
	int N, wpl, weight[MAXN];
	scanf("%d\n", &N);
	M = Create(N);
	Establish(M,Orig);
	H = Huffman(M);
	wpl = WPL(H, 0);
	Check(wpl, N, Orig);
	system("pause");
	return 0;
}



void Insert(PtrM M, PtrH item)
{
	int i;
	i = ++M->size;
	for (; M->Elems[i / 2]->wei >= item->wei; i /= 2)
		M->Elems[i] = M->Elems[i / 2];
	M->Elems[i] = item;
}
PtrH Delete(PtrM M)
{
	int Parent, Child;
	PtrH Minitem, temp;
	Minitem = M->Elems[1];
	temp = M->Elems[M->size--];
	for (Parent = 1; Parent * 2 <= M->size; Parent = Child)
	{
		Child = Parent * 2;
		if ((Child != M->size) && (M->Elems[Child]->wei > M->Elems[Child + 1]->wei))
			Child++;
		if (temp->wei <= M->Elems[Child]->wei)
			break;
		else M->Elems[Parent] = M->Elems[Child];
	}
	M->Elems[Parent] = temp;
	return Minitem;
}
PtrM Create(int maxsize)
{
	PtrM M;
	M = (PtrM)malloc(sizeof(struct Min_Heap));
	M->cap = maxsize;
	M->Elems[0] = (PtrH)malloc(sizeof(struct HaffCode));
	M->size = 0;
	M->Elems[0]->wei = -1;
	M->Elems[0]->rep = -1;
	M->Elems[0]->left = M->Elems[0]->right = NULL;
	return M;
}
void Establish(PtrM M,PtrO Orig)
{
	int i;
	M->size = M->cap;
	for (i = 1; i size; i++)
	{
		M->Elems[i] = (PtrH)malloc(sizeof(struct HaffCode));
		scanf("%c%d ", &(M->Elems[i]->rep),&(M->Elems[i]->wei));
		Orig[i-1].rep = M->Elems[i]->rep;
		Orig[i-1].weight = M->Elems[i]->wei;
		M->Elems[i]->right = M->Elems[i]->left = NULL;
	}
	M->Elems[i] = (PtrH)malloc(sizeof(struct HaffCode));
	scanf("%c%d", &(M->Elems[i]->rep),&(M->Elems[i]->wei));
	Orig[i - 1].rep = M->Elems[i]->rep;
	Orig[i - 1].weight = M->Elems[i]->wei;
	M->Elems[i]->right = M->Elems[i]->left = NULL;
	for ( i = M->size / 2; i > 0; i--)
		PercDown(M, i);
}
void PercDown(PtrM M, int index)
{
	int Parent, Child;
	PtrH H = M->Elems[index];
	for (Parent = index; Parent * 2 <= M->size; Parent = Child)
	{
		Child = Parent * 2;;
		if ((Child!=M->size)&&(M->Elems[Child]->wei > M->Elems[Child + 1]->wei))
			Child++;
		if (H->wei <= M->Elems[Child]->wei)
			break;
		else
			M->Elems[Parent] = M->Elems[Child];
	}
	M->Elems[Parent] = H;
}
PtrH Huffman(PtrM M)
{
	int i;
	PtrH H;
	for (i = 1; M->size>1; i++)
	{
		H = (PtrH)malloc(sizeof(struct HaffCode));
		H->left = Delete(M);
		H->right = Delete(M);
		H->wei = H->left->wei + H->right->wei;
		H->rep = -1;
		Insert(M, H);
	}
	H = Delete(M);
	return H;
}
int WPL(PtrH H, int height)
{
	if (H->rep != -1)
		return H->wei*height;
	else return WPL(H->left, height + 1) + WPL(H->right, height + 1);
}
void Print(PtrH H)
{
	if (H)
	{
		if (H->rep != -1)
			printf("%c %d ", H->rep, H->wei);
		Print(H->left);
		Print(H->right);
	}
}
void Check(int WPL, int N, PtrO Orig)
{
	int M, SubW;
	Submit Sub[MAXN];
	scanf("%d", &M);
	for (int i = 0; i < M; i++)
	{
		SubW = 0;
		for (int j = 0; j < N; j++)
		{
			scanf("\n%c", &Sub[j].rep);
			scanf(" %s", Sub[j].code);
			Sub[j].length = strlen(Sub[j].code);
			SubW += FindWeight(Orig, Sub[j].rep, N)*Sub[j].length;
		}
		if (SubW > WPL)
		{
			printf("No\n");
			continue;
		}
		int k;
		PtrJ J = NewNode();
		for (k = 0; k < N; k++)
			if (!Judge(Sub[k].code, J))
				break;
		if (k == N)
			printf("Yes\n");
		else printf("No\n");
	}
}
int FindWeight(PtrO Orig, char rep,int N)
{
	int i;
	for (i = 0; i < N; i++)
		if (Orig[i].rep == rep)
			return Orig[i].weight;
	printf("No found\n");
	return -1;
}
PtrJ NewNode(void)
{
	PtrJ J;
	J = (PtrJ)malloc(sizeof(struct JNode));
	J->flag = 0;
	J->left = NULL;
	J->right = NULL;
	return J;
}
int Judge(char *s, PtrJ J)
{
	int i, len;
	len = strlen(s);
	for (i = 0; i < len; i++)
	{
		if (s[i] == '0')
		{
			if (!J->left)
				J->left = NewNode();
			else if (J->left->flag)
				return 0;
			J = J->left;
		}
		else
		{
			if (!J->right)
				J->right = NewNode();
			else if (J->right->flag)
				return 0;
			J = J->right;
		}
	}
	J->flag = 1;
	if ((!J->left) && (!J->right))
		return 1;
	else return 0;
}

写的又臭又长…