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:
[code]c[1] f[1] c[2] f[2] ... c f
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:
[code]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:
[code]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:
[code]Yes Yes No No
[code]/* 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; }
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