PTA_2019春_059_ 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 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

/* Huffman Codes*/
/*思路：将输入的每个字符建树，若每个数为叶节点（叶节点可使其不出现二义性），且满足最短编码，为正确编码 
		最短编码可由哈夫曼树获得
		不需要判断为叶节点，直接判断一个字母是否是另一个字母的前缀来避免二义性*/
		/*总感觉树不应该用链式来建，后序操作太过麻烦（遍历，从下往上，找值等等操作），考虑一下用结构体数组来*/
		/*5/16/19： 判断是否为前缀码时没有考虑两个码一样*/
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
typedef struct data* Data;  
struct data {
	char a;
	int num;
};
typedef struct studentdata* Sdata;  /*学生输入数据结构*/ 
struct studentdata {
	char a;
	char array[70];

};
typedef struct HFnode* HFtree;
struct HFnode {   /*哈夫曼树定义*/
	int weight;  /*权值*/
	char data;
	HFtree left;
	HFtree right;
	HFtree parent;  /*方便回溯*/
};
typedef struct Hnode* Heap;/*堆的定义*/
struct Hnode {
	HFtree* data;
	int size;
	int capacity;
};
typedef struct Snode* Stack; /*堆栈*/
struct Snode {
	HFtree* data;
	int top;
	int maxsize;
};

void insert(Heap H, HFtree tree);   /*最小堆插入*/ 
HFtree deletemin(Heap H);  /*堆删除*/ 
void traversal(HFtree T);  /*测试用的遍历*/ 
Stack creatstack(int maxsize);  /*创建栈*/ 
void push(Stack S, HFtree T); /*进栈*/ 
HFtree pop(Stack S);  /*出栈*/ 
int isempty(Stack S); /*是否为空*/ 
HFtree find(HFtree T, char a);  /*在哈弗树中找节点*/ 
int WPL(HFtree T, Data array, int N); /*计算wpl*/ 
void isture(int wpl, int N, Data indata); /*判断是否符合题意*/ 
int ispre(int N, Sdata sdata);  /*判断是否为前缀编码*/ 
int max(int a, int b);  /*获得两者较大值*/ 

int main() {
	int N;/*节点个数*/
	char laji;/*消除\n 和空格*/
	scanf("%d", &N);
	Data indata = (Data)malloc(N * sizeof(struct data));
	for (int i = 0; i < N; i++) {
		scanf("%c", &laji);
		scanf("%c", &indata[i].a);
		scanf("%d", &indata[i].num);
	}
	
	/*创建最小堆，并将输入的数据存在最小堆中，留着建树用*/
	Heap minheap = (Heap)malloc(sizeof(struct Hnode));
	minheap->data = (HFtree*)malloc((N + 1) * sizeof(HFtree));    /*下面几行给我往死了里理解，想想自己为什么一开始是错的*/
	for (int i = 0; i < N + 1; i++) {     
		minheap->data[i] = (HFtree)malloc(sizeof(struct HFnode));
	}

	minheap->size = 0;
	minheap->capacity = N;
	minheap->data[0]->weight = -2;
	minheap->data[0]->left = minheap->data[0]->right = minheap->data[0]->parent = NULL;

	for (int i = 0; i < N; i++) {
		HFtree hftree = (HFtree)malloc(sizeof(struct HFnode));
		hftree->data = indata[i].a;
		hftree->weight = indata[i].num;
		hftree->left = hftree->right = hftree->parent = NULL;
		insert(minheap, hftree);
	}/*最小堆建成*/

	/*建树*/
	HFtree T;
	for (int i = 1; i < N; i++) {
		T = (HFtree)malloc(sizeof(struct HFnode));
		T->parent = NULL;
		T->left = deletemin(minheap);
		T->left->parent = T;
		T->right = deletemin(minheap);
		T->right->parent = T;
		T->weight = T->left->weight + T->right->weight;
		insert(minheap, T);
	}
	T = deletemin(minheap);/*堆中最后一个元素即树的根节点*/
	//traversal(T);

	/*获得最小编码数*/
	/*测试find函数*/
	int wpl;       /*最小编码数*/
	wpl = WPL(T, indata, N);
	/*测试*/
//	printf("wpl is %d\n", wpl);
	/*处理学生的输入数据*/
	int L;/*学生个数*/
	scanf("%d", &L);
	scanf("%c", &laji);
	for (int j = 0; j < L; j++) {
		isture(wpl, N, indata);
	}
	return 0;
}
void insert(Heap H, HFtree tree) {
	int i;
	H->size++;
	i = H->size; /*i指向将要插入的位置*/
	for (; H->data[i / 2]->weight > tree->weight; i /= 2) {
		H->data[i] = H->data[i / 2];
	}
	H->data[i] = tree;

}
HFtree deletemin(Heap H) {
	int parent, child;
	HFtree min, x;
	min = H->data[1];/*先取出最小值*/
	/*下面重新调整*/
	x = H->data[H->size];
	H->size--;
	for (parent = 1; parent * 2 <= H->size; parent = child) {
		child = 2 * parent;
		if ((child != H->size) && (H->data[child]->weight > H->data[child + 1]->weight)) {
			child++;
		}
		if (x->weight <= H->data[child]->weight) break;
		else {
			H->data[parent] = H->data[child];
		}
	}
	H->data[parent] = x;
	return min;
}
void traversal(HFtree T) {
	if (T) {
		traversal(T->left);
		traversal(T->right);
		if (!T->left && !T->right) {

			printf("%c\n", T->data);
		}
	}
}
Stack creatstack(int maxsize) {
	Stack S = (Stack)malloc(sizeof(struct Snode));
	S->data = (HFtree*)malloc(maxsize * (sizeof(HFtree)));  /*此处同上面那个重要的错误*/
	S->maxsize = maxsize;
	S->top = -1;
	return S;
}
void push(Stack S, HFtree T) {
	S->data[++(S->top)] = T;
}
HFtree pop(Stack S) {
	return(S->data[(S->top)--]);
}
int isempty(Stack S) {
	return(S->top == -1);
}
HFtree find(HFtree T, char a) {
	HFtree tree;
	Stack S = creatstack(500);
	tree = T;
	while (tree || isempty(S) != 1) {
		while (tree) {
			push(S, tree);
			tree = tree->left;
		}
		tree = pop(S);
		if (!tree->left && !tree->right) {
			if (tree->data == a) return tree;
		}
		tree=tree->right;
	}
}
int WPL(HFtree T, Data array, int N) {
	int cnt = 0;
	int weight;
	HFtree tree;
	for (int i = 0; i < N; i++) {
		int k = 0;  /*路径长度*/
		tree = find(T, array[i].a);
		weight = tree->weight;
		while (1) {
			if (tree == T) break;
			tree = tree->parent;
			k++;
		}
		cnt = cnt + k * weight;
	}
	return cnt;
}
void isture(int wpl, int N, Data indata) {
	int student_wpl = 0;
	char laji;
	Sdata sdata = (Sdata)malloc(N * sizeof(struct studentdata));
	for (int i = 0; i < N; i++) {
		scanf("%c", &sdata[i].a);
		scanf("%c", &laji);
		gets(sdata[i].array);
	}
	/*测试,sdata数组没有问题*/
	for (int i = 0; i < N; i++) {
		student_wpl = student_wpl + indata[i].num * strlen(sdata[i].array);/*默认两数组的元素对应*/
	}
	if (wpl == student_wpl) {
		if (ispre(N, sdata))  printf("Yes\n");
		else printf("No\n");
	}
	else printf("No\n");

}
int ispre(int N, Sdata sdata) {/*两个字符串一个一个进行对比,当不同时，若此时其中一个字符串已经结束，那么不符合前缀*/
	for (int i = 0; i < N-1; i++) {
		for (int j = i + 1; j < N; j++) {
			for (int k = 0;k<70; k++) {
				if (sdata[i].array[k] != sdata[j].array[k]) {      /*当两个字符串比较到不相等时，判断此时是否有字符串遍历结束，若有结束则不为前缀*/ 
					if (strlen(sdata[i].array) == k || strlen(sdata[j].array) == k) {
						return 0;
					}
					break;
				}
				if(k==68)  return 0;     /*两个编码一样*/
			}
		}
	}
	return 1;
}
int max(int a, int b) {
	if (a > b)  return a;
	else return b;
}