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), 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;

#define maxn 70

int N,codelen,cnt1,cnt2,w[maxn];

char ch[maxn];

typedef struct TreeNode* Tree;

struct TreeNode{

    int weight;

    Tree Left,Right;

};

typedef struct HeapNode* Heap;

struct HeapNode{

    struct TreeNode Data[maxn];

    int size;

};

Tree creatTree(){

      Tree T;

      T = new struct TreeNode;

      T->weight = ;

      T->Left = T->Right = NULL;

      return T;

}

Heap creatHeap(){

    Heap H;

    H = new struct HeapNode;

    H->Data[].weight = -;

    H->size = ;

    return H;

}

void Insert(Heap H,struct TreeNode T){

    int i = ++H->size;

    for(; T.weight < H->Data[i/].weight; i /= )

         H->Data[i] = H->Data[i/];

    H->Data[i] = T;

}

Tree Delete(Heap H){

    int child,parent;

    struct TreeNode Temp = H->Data[H->size--];

    Tree T = creatTree();

    *T = H->Data[];

    for(parent = ;  * parent <= H->size; parent = child){

        child =  * parent;

        if(child < H->size && H->Data[child].weight > H->Data[child+].weight)

        child++;

        if(H->Data[child].weight > Temp.weight) break;

        H->Data[parent] = H->Data[child];

    }

    H->Data[parent] = Temp;

    return T;

}

Tree Huffman(Heap H){

    Tree T = creatTree();

    while(H->size != ){

        T->Left = Delete(H);

        T->Right = Delete(H);

        T->weight = T->Right->weight + T->Right->weight;

        Insert(H,*T);

    }

    T = Delete(H);

    return T;

}

int WPL(Tree T,int depth){

    if(!T->Left && !T->Right) return(depth*T->weight);

    else return WPL(T->Left,depth+)+WPL(T->Right,depth+);

}

void JudgeTree(Tree T){

    if(T){

        if(T->Right && T->Left) cnt2++;

        else if(!T->Left && !T->Right) cnt1++;

        else cnt1 = ;

        JudgeTree(T->Left);

        JudgeTree(T->Right);

    }

}

int Judge(){

    int i,j,wgh,flag = ;;

    char s1[maxn],s2[maxn];

    Tree T = creatTree(), pt = NULL;

    for(i = ; i < N; i++){

        cin >> s1 >> s2;

        if(strlen(s2) > N) return ;

        for(j = ; s1[] != ch[j]; j++); wgh = w[j];

        pt = T;

        for(j = ; s2[j] ; j++){

            if(s2[j] == ''){

                if(!pt->Left) pt->Left = creatTree();

                pt = pt->Left;

            }

            if(s2[j] == ''){

                if(!pt->Right) pt->Right = creatTree();

                pt = pt->Right;

            }

            if(pt->weight) flag = ;

            if(!s2[j+]){

                if(pt->Left || pt->Right) flag = ;

                else pt->weight = wgh;

            }

        }

    }

    if(flag == ) return ;

    cnt1 = cnt2 = ;

    JudgeTree(T);

    if(cnt1 != cnt2 + ) return ;

    if(codelen == WPL(T,)) return ;

    else return ;

}

int main(){

    int i,n;

    Tree T;

    Heap H;

    T = creatTree();

    H = creatHeap();

    cin >> N;

    for(i = ; 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,);

    cin >> n;

    while(n--){

        if(Judge()) cout<< "Yes" << endl;

        else cout << "No" << endl;

    }

    return ;

}

巴特西

05-树9 Huffman Codes （30 分）

Input Specification:

Output Specification:

Sample Input:

Sample Output:

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