7-9 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≤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
Code:
#include <bits/stdc++.h>
using namespace std;struct node
{int pro;bool operator<(const node &a) const{return pro > a.pro;}
};priority_queue<node> p;
map<char, int> ctr;
int sum = 0;
int n, m;bool cmp2(string a, string b)
{return a.length() > b.length();
}int main()
{cin >> n;for (int i = 0; i < n; i++){char a;int b;cin >> a >> b;// ctr[a] = b;ctr.insert(pair<char, int>(a, b));node n1;n1.pro = b;p.push(n1);}// 一共n-1次合并 计算valuefor (int i = n; i < 2 * n - 1; i++){node n1, n2, n3;n1 = p.top();p.pop();n2 = p.top();p.pop();n3.pro = n1.pro + n2.pro;sum += n3.pro;p.push(n3);}cin >> m;for (int i = 0; i < m; i++){int sum2 = 0;string s[1000];for (int j = 0; j < n; j++){char a;string b;cin >> a >> b;sum2 += ctr[a] * b.length();s[j] = b;}if (sum != sum2){cout << "No" << endl;continue;} //判权值和是否相等sort(s, s + n - 1, cmp2);bool flag = 1;for (int x = 0; x < n - 1; x++)for (int y = x + 1; y < n; y++)if (s[x].find(s[y]) == 0)flag = 0; //任意编码不能是其余编码的前缀if (flag)cout << "Yes" << endl;elsecout << "No" << endl;}return 0;
}