7-9 Huffman Codes (30分) 哈夫曼编码

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

虽然题目说是哈夫曼编码，实际只要权值和是最小，编码中不存在某个编码是另一个编码的前缀，且编码长度符合条件(比n小)就可以。
首先用优先队列根据频率计算出最小的权值和，接着对于每个学上提交的编码，用字典树判断是否存在前缀重合的问题，同时根据编码长度和频率计算权值和然后比较，而且判断每个编码长度要合格。

代码：

#include <cstdio>
#include <cstring>
#include <queue>
using namespace std;
int n,rc,c,d,num[300],k,flag;
char s[2],t[100];
int trie[10000][2],ind,en[10000];
void build(char *str) {
    int r = 0,i = 0;
    while(str[i]) {
        int e = str[i ++] - '0';
        if(!trie[r][e]) {
            trie[r][e] = ++ ind;
        }
        r = trie[r][e];
        if(en[r] == 1) flag = 1;
        if(str[i]) en[r] = 2;
    }
    if(en[r]) flag = 1;
    en[r] = 1;
}
int main() {
    scanf("%d",&n);
    priority_queue<int,vector<int>,greater<int> > q;
    for(int i = 0;i < n;i ++) {
        scanf("%s %d",s,&d);
        num[(int)s[0]] = d;
        q.push(d);
    }
    while(q.size() > 1) {
        d = q.top();
        q.pop();
        d += q.top();
        q.pop();
        q.push(d);
        rc += d;
    }
    scanf("%d",&k);
    for(int i = 0;i < k;i ++) {
        c = ind = flag = 0;
        memset(trie,0,sizeof(trie));
        memset(en,0,sizeof(en));
        for(int j = 0;j < n;j ++) {
            scanf("%s %s",s,t);
            int len = strlen(t);
            if(len >= n) flag = 1;
            if(flag) continue;
            build(t);
            c += num[int(s[0])] * len;
        }
        printf("%s
",c == rc && !flag ? "Yes" : "No");
    }
}