05-树9 Huffman Codes (30分)

2023-12-09 12:08

文章标签 05 30 huffman codes

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05-树9 Huffman Codes (30分)

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

#include<bits/stdc++.h>
using namespace std;
int Check(string ss[],int N);
int main(){map<char,int>m;priority_queue<int, vector<int>,greater<int> >q;int N,V,num,val,len;char c;cin>>N;for(int i=0;i<N;i++){cin>>c>>num;m[c]=num;q.push(num);}while(1){num=q.top();q.pop();if(q.empty())break;num+=q.top();q.pop();q.push(num);val+=num;}cin>>V;for(int i=0;i<V;i++){len=0;char t;string ss[N];for(int i=0;i<N;i++){cin>>t>>ss[i];int l=ss[i].size();len+=l*m[t];}if(len!=val||!Check(ss,N))printf("No\n");else printf("Yes\n");}
}
int Check(string ss[],int N){sort(ss,ss+N);for(int i=0;i<N-1;i++)for(int j=i+1;j<N;j++)if(ss[j].find(ss[i])==0)return 0;return 1;
}

#include<bits/stdc++.h>
using namespace std;typedef struct TNode* Ptr;
struct TNode {char c;int f;
};int N, Size;
Ptr Heap[100];
map<char, int> codeMap;void Insert();
Ptr Delete();
void Pre(int root);int main() {cin >> N;char c;int fre, sum = 0;Size  = 0;for (int i = 1; i<= N; i++) {cin >> c >> fre;codeMap[c] = fre;getchar();Heap[i] = (Ptr)malloc(sizeof(struct TNode));Heap[i]->c = c;Heap[i]->f = fre;Size++;Insert();}for (int i = 1; i < N; i++) {Ptr root1 = Delete();Ptr root2 = Delete();root1->f += root2->f;sum += root1->f;Heap[++Size] = root1;free(root2);}int number;cin>>number;while (number--) {char C1;map<string, bool> fuck;string code[N], s1, s2;int judge = 0, flag = 0;for (int i = 0; i < N; i++) {cin >> C1 >> code[i];if (flag == 1) continue;if (fuck[code[i]] == true) flag = 1;fuck[code[i]] = true;getchar();judge += (codeMap[C1] * code[i].length());}for (int i = 0; i < N; i++) {code[i] = code[i].substr(0, code[i].length()-1);if (fuck[code[i]] == true) flag = 1;}if (judge == sum && !flag) printf("Yes\n");else printf("No\n");}
}void Insert() {Ptr Temp = Heap[Size];int X = Heap[Size]->f, i = Size;for (; i > 1 && Heap[i / 2]->f > X; i /= 2)Heap[i] = Heap[i / 2];Heap[i] = Temp; 
}Ptr Delete() {if (!Heap[1]) return nullptr;Ptr Temp = Heap[1];Heap[1] = Heap[Size--];Pre(1);return Temp; 
}void Pre(int root) {Ptr Temp = Heap[root];int X = Heap[root]->f, Parent = root, Child;for (; Parent * 2 <= Size; Parent = Child) {Child = Parent * 2;if (Child != Size && Heap[Child]->f > Heap[Child + 1]->f)Child++;if (X <= Heap[Child]->f)break;else Heap[Parent] = Heap[Child]; }Heap[Parent] = Temp;
}