7-9 Huffman Codes (30 分)- C语言实现

太心累了，主要是一开始建堆出现了小问题，导致测试点一直没过。心烦意乱，采用printf调试（bushi），终于找出来了，是因为child的范围没考虑，忘记限制2*i

简单记录下要点，吃饭去了：

哈夫曼编码要点

1. 编码长度之和一定等于哈夫曼编码的长度之和，也就是哈夫曼树各个叶结点的路径之和，所以采用哈夫曼树主要就是为了算出路径长度，图省事，我也没建立树，直接算总和。

2. 要满足前缀码都不相同，cyll在mooc上采用的是构建树的方法去看是否有字符落在不是叶节点上，我偷懒直接采用strcmp暴力遍历

满足这两点就可以算得上是最优编码了。

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

结尾无空行

Sample Output:

Yes
Yes
No
No

结尾无空行

解答

#include 
#include 
#include 
//如果用struct的话可以真正实现哈夫曼树
//但这道题目其实只要求我们求出哈夫曼编码的长度，所以可以采用整型变量作数组类型
int N;
int char_weight[255];
int cnt;//the size of number
typedef struct Node* PtrNode;
typedef struct Node{
    int data;
    PtrNode left;
    PtrNode right;
}Node;
typedef PtrNode Tree;
void Precdown(int heap[],int p){
    int tmp=heap[p];
    int i,child;
    for(i=p;2*i<=N;i=child){
        child=2*i;
        if(child1;i/=2){
        if(heap[i/2]>tmp){
            heap[i]=heap[i/2];
        }else break;
    }
    heap[i]=tmp;
}
int  DeleteHeap(int heap[]){
    int min=heap[1];
    heap[1]=heap[N];
    N--;
    Precdown(heap,1);
    return min;
}
void InsertHeap(int heap[],int x){
    N++;
    heap[N]=x;
    Precup(heap,N);
}
void CreateMinHeap(int heap[]){

    int i;
    for(i=N/2;i>0;i--){
        Precdown(heap,i);
    }
}
int BuildHuff(int heap[]){
    int length=0;
    int v1,v2;
    int root;
    while(N>1){
        v1=DeleteHeap(heap);
        v2=DeleteHeap(heap);
        root=v1+v2;
        length+=root;
        InsertHeap(heap,root);
    }
    return length;
}
int Judge(int length){
    char c[100][100];
    int total=0;
    int num[255]={0};
    int i=0,j=0;
    int slength;
    char ch;
    int l1,l2;
    for(i=0;i 
比较混乱，有闲情逸致再来修改总结。挥挥~