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;
const int maxn = 1010;

int n,m;
int wgh;
int cnt1,cnt2;
int w[maxn];
char ch[maxn];
int codelen;

typedef struct TreeNode* Tree;
struct TreeNode{
    int weight;
    Tree Left,Right;
};
typedef struct HeapNode* Heap;
struct HeapNode{
    TreeNode Data[maxn];
    int size;
};

Tree creatTree();
Heap creatHeap();
void Insert(Heap H,TreeNode T);
Tree Delete(Heap H);
Tree Huffman(Heap H);
int WPL(Tree T,int depth);
void JudgeTree(Tree T);
bool Judge();

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);
     cin >> m;
     while(m--){
         if(Judge()) printf("Yes
");
         else printf("No
");
     }
     
    return 0;
}

Tree creatTree(){
    Tree T = new TreeNode;
    T->weight = 0;
    T->Left = T->Right = NULL;
    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(;T.weight < H->Data[i/2].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 child,parent;
    TreeNode temp = H->Data[H->size--];
    Tree T = creatTree();
    *T = H->Data[1];
    for(parent = 1; 2 * parent <= H->size; parent = child){
        child = parent * 2;
        if(child < H->size && H->Data[child+1].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;
}

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;
    for(int i = 0; i < n; i++){
        cin >> s1 >> s2;
        if(strlen(s2) > n) return 0;
        int j;
        for(j = 0; s1[0] != ch[j];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;
    cnt1 = cnt2 = 0;
    JudgeTree(T);
    if(cnt1 != cnt2 + 1) return 0;
    if(codelen == WPL(T,0)) return 1;
    else return 0;
}

void JudgeTree(Tree T){
    if(T){
        if(T->Right&&T->Left) cnt2++;
        else if(!T->Right && !T->Left) cnt1++;
        else cnt1 = 0;
        JudgeTree(T->Left);
        JudgeTree(T->Right);
    }
}