05-树9 Huffman Codes

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 NN (2\le N\le 632≤N≤63), then followed by a line that contains all the NN 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 MM (\le 1000≤1000), then followed by MM student submissions. Each student submission consists of NN 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

题目大意：判断给出的哈夫曼编码是否正确

思路：不用重新构建哈夫曼编码，因为可能有数种情况，只要比较其花费是否最小，编码是否出现前序重复就可以了。

#include<iostream>#include<functional>#include <string>  #include <algorithm>  #include <queue>  using namespace std;typedef struct{    char alpha;    int frequency;}inputdata;typedef struct{    char alpha;    string code;}check;int getfrequency(char alpha,inputdata* data,int num){    for (int i = 0; i < num; i++)    {        if (data[i].alpha == alpha)            return data[i].frequency;    }    return 0;}bool cmp(check ch1, check ch2){    return ch1.code.size() < ch2.code.size();}bool isoverlap(check* ch,int num){    sort(ch, ch + num, cmp);    for (int i = 0; i < num; i++)    {        string tmp = (ch + i)->code;        for (int j = i + 1; j < num; j++)        {            if ((ch + j)->code.substr(0, tmp.size()) == tmp)                return true;        }    }    return false;}int main(){    int num;    cin >> num;    int leastcost = 0;    inputdata* input = new inputdata[num];    priority_queue<int, vector<int>, greater<int>> q;    for (int i = 0; i < num; i++)    {        cin >> (input + i)->alpha;        cin >> (input + i)->frequency;        q.push((input + i)->frequency);    }    int sum;    while (!q.empty())    {        sum = 0;        sum += q.top();        q.pop();        if (!q.empty())        {            sum += q.top();            q.pop();            q.push(sum);        }        leastcost += sum;    }    int N;    cin >> N;    int cost;    int* flag = new int[N];    check* temp = new check[num];    for (int i = 0; i < N; i++)    {        cost = 0;        for (int j = 0; j < num; j++)        {            cin >> (temp + j)->alpha;            cin >> (temp + j)->code;        }        for (int j = 0; j < num; j++)        {            cost += (temp[j].code.size()+1)*getfrequency(temp[j].alpha, input, num);        }        if (cost>leastcost)        {            //flag[i] = 0;            printf("No\n");        }        else        {            if (isoverlap(temp, num))                //flag[i] = 0;                printf("No\n");            else                //flag[i] = 1;                printf("Yes\n");        }    }    delete[]flag;    delete[]input;    delete[]temp;    return 0;}