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 (≤), 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
#include<iostream> #include<cstring> using namespace std; const int maxn = 1010; typedef struct TreeNode* Tree; struct TreeNode { Tree left,right; int weight; }; typedef struct HeapNode* Heap; struct HeapNode { TreeNode Data[maxn]; int size; }; int n,m; int w[maxn]; char ch[maxn]; int codelen; int cnt2,cnt0; Tree creatTree(); Heap creatHeap(); void Insert(Heap H, TreeNode T); Tree Huffman(Heap H); Tree Delete(Heap H); int WPL(Tree T, int depth); bool Judge(); void JudgeTree(Tree T); int main() { cin >> n; Tree T = creatTree(); Heap H = creatHeap(); for (int i = 0; i < n; i++) { getchar(); cin >> ch[i] >> w[i]; H->Data[H->size].left = H->Data[H->size].right = NULL; T->weight = w[i]; Insert(H,*T); } T = Huffman(H); codelen = WPL(T,0); //printf("%d ",codelen); cin >> m; while (m--) { if (Judge()) { printf("Yes "); } else { printf("No "); } } return 0; } Tree creatTree() { Tree T = new TreeNode; T->left = T->right = NULL; T->weight = 0; return T; } Heap creatHeap() { Heap H = new HeapNode; H->Data[0].weight = -1; H->size = 0; return H; } void Insert(Heap H, TreeNode T) { int i = ++H->size; for (; H->Data[i/2].weight > T.weight; i /= 2) { H->Data[i] = H->Data[i/2]; } H->Data[i] = T; } Tree Huffman(Heap H) { Tree T = creatTree(); while (H->size > 1) { T->left = Delete(H); T->right = Delete(H); T->weight = T->left->weight + T->right->weight; Insert(H,*T); } T = Delete(H); return T; } Tree Delete(Heap H) { int parent,child; TreeNode Tmp = H->Data[H->size--]; Tree T = creatTree(); *T = H->Data[1]; for (parent = 1; 2*parent <= H->size; parent = child) { child = 2*parent; if (child < H->size && H->Data[child+1].weight < H->Data[child].weight) { child++; } if (H->Data[child].weight > Tmp.weight) { break; } H->Data[parent] = H->Data[child]; } H->Data[parent] = Tmp; return T; } int WPL(Tree T, int depth) { if (!T->left && !T->right) { return depth * (T->weight); } else { return WPL(T->left,depth+1) + WPL(T->right,depth+1); } } bool Judge() { char s1[maxn],s2[maxn]; bool flag = true; Tree T = creatTree(); Tree pt = NULL; int wgh; for (int i = 0; i < n; i++) { cin >> s1 >> s2; if (strlen(s2) > n) { return 0; } int j; for (j = 0; ch[j] != s1[0]; j++) { ; } wgh = w[j]; pt = T; for (j = 0; s2[j]; j++) { if (s2[j] == '0') { if (!pt->left) { pt->left = creatTree(); } pt = pt->left; } if (s2[j] == '1') { if (!pt->right) { pt->right = creatTree(); } pt = pt->right; } if (pt->weight) { flag = false; } if (!s2[j+1]) { if (pt->left || pt->right) { flag = false; } else { pt->weight = wgh; } } } } if (!flag) { return 0; } cnt0 = cnt2 = 0; JudgeTree(T); if (cnt2 != cnt0-1) { return 0; } if (codelen == WPL(T,0)) { return 1; } else { return 0; } } void JudgeTree(Tree T) { if (T) { if (!T->left && !T->right) { cnt0++; } else if(T->left && T->right) { cnt2++; } else { cnt0 = 0; } JudgeTree(T->left); JudgeTree(T->right); } }