标签：and break follow pmi heap node Fix bit main
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.
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 Ndistinct 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.
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.
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 11Yes
Yes
No
No
参考文章 https://zhuanlan.zhihu.com/p/121684742
借助小顶堆建立huffmanTree ,计算WPL
然后用学生输入的huffmanCode 建立huffmanTree,
建树过程中,累加 (对应字符的频率*哈夫曼码长度) 得到学生的树的带权路径长度wplOfTest
如果结点存放的位置 非叶结点 或 已经存放了其他结点信息 记为错误huffmanCode, 跳过其他字符的哈夫曼码的处理
如果学生的huffmanCode可以生成huffmanTree, 则计算学生的带权路径长度 即wplOfTest与WPL是否相等, 不相等也是错误的huffmanCode
//小顶堆的建立 查找 #include <iostream> #include <vector> using namespace std; class tnode{ public: string c{""}; int f{0}; tnode* left{nullptr}; tnode* right{nullptr}; tnode()=default; tnode(string c_,int f_):c{c_},f{f_}{}; }; class minHeap{//小根堆 public: vector<tnode*> heap; minHeap(){ } int getSize(){ return heap.size(); } void build(vector<tnode*> list,int n){ for(int i=0;i<n;i++){ insertNode(list[i]); } } void insertNode(tnode* newnode){ heap.push_back(newnode); adjustFromBack(); } void insertNode(string c,int f){ tnode* newnode =new tnode{c,f}; heap.push_back(newnode); adjustFromBack(); } tnode* popMinNode(){ tnode* temp{nullptr}; if(heap.size()){ temp = heap.front(); swap(heap.front(), heap.back()); heap.pop_back(); adjustFromFront(); } return temp; } void adjustFromBack(){ for(int i=getSize()-1;i>=0&&getSize()>1;i--){ if(heap[i]->f<heap[(i-1)/2]->f){ swap(heap[i], heap[(i-1)/2]); } } } void adjustFromFront(){ for(int i=1;i<getSize()&&getSize()>1;i++){ if(heap[i]->f<heap[(i-1)/2]->f){ swap(heap[i], heap[(i-1)/2]); } } } }; class huffmanTree{ public: huffmanTree()=default; tnode* root; void create(minHeap &minheap){ tnode* n1,*n2,*newnode; int size=minheap.getSize(); for(int i=0;i<size-1;i++){ n1 = minheap.popMinNode(); n2 = minheap.popMinNode(); newnode = new tnode{"",n1->f+n2->f}; newnode->left = n1; newnode->right = n2; minheap.insertNode(newnode); } newnode = minheap.heap.front(); root=newnode; } int WPL(tnode* p,int depth){ if(!p->left&&!p->right) return p->f*depth; return WPL(p->left, depth+1) + WPL(p->right, depth+1); } int getWPL(){ return WPL(root, 0); } }; bool judge(int wpl,vector<tnode*>list){ bool flag=true; int wplOfTest{0}; string c,code; tnode* head=new tnode; tnode* tail=head; tnode* temp; for(int i=0;i<list.size();i++){ cin >> c >> code; if(!flag)continue;; wplOfTest+=(list[i]->f*code.size()); for(auto i=code.begin();i!=code.end();i++){ if(*i==‘0‘){ if(!tail->left){ temp = new tnode; tail->left=temp; } tail=tail->left; }else if(*i==‘1‘){ if(!tail->right){ temp=new tnode; tail->right=temp; } tail=tail->right; } if(tail->f){ flag=false; break; } } if(tail->left||tail->right){ flag=false; continue; } tail->f=list[i]->f; tail->c=list[i]->c; tail=head; } if(wplOfTest!=wpl)return false; else return flag; } int main(){ int n,m,WPL; string c; int f; cin >> n; vector<tnode*> alpha; for(int i=0;i<n;i++){ cin >> c >> f; tnode *newnode=new tnode{c,f}; alpha.push_back(newnode); } minHeap minheap; minheap.build(alpha,n); huffmanTree ht; ht.create(minheap); WPL=ht.getWPL(); cin >> m; for(int i=0;i<m;i++){ if(judge(WPL,alpha)){ cout << "Yes"<<endl; }else{ cout << "No"<<endl; } } return 0; }
数据结构 05-树9 Huffman Codes (30 分)
标签：and break follow pmi heap node Fix bit main
原文地址：https://www.cnblogs.com/ichiha/p/14788139.html
