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 Mstudent 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思路：
1、首先自己根据这些字符和频率建立哈夫曼树，求出最小WPL。
2、计算出每个学生编码的WPL，即用字符串的长度乘以频率。
3、用String中的startwith（）函数两两判断一个字符串是否是另一个字符串的前缀。当满足前缀码要求并且WPL与哈夫曼计算出的WPL相等时，认为是正确的。
4、建哈夫曼树时，可以利用最小堆（优先队列）输出2个最小的。合并之后再添加到最小堆中。我是直接给数组排序了，没用最小堆。
5、整体思路比较简单，有些坑容易跳，比如，学生输入的字符顺序可能是乱的，甚至字符输入的就不是给定的字符？有两个测试点到现在还是通不过，等9月份再次开课之后再进去试试，看看到底是什么测试点。
import java.util.Arrays;
import java.util.Scanner;
public class Main {
private static int sum=0;
public static void main(String[] args) {
// TODO Auto-generated method stub
Scanner in = new Scanner(System.in);
int n = in.nextInt(); //记录总的字符数量
char [] ch = new char[n];
int [] fr = new int[n];
for (int i = 0;i<n;i++) {
ch[i]=in.next().charAt(0);
fr[i]=in.nextInt();
}
int stu =in.nextInt();
int wpl_min = huffman_code_wpl(fr);
for(int i = 0;i<stu;i++) {
judge(in,ch,fr,wpl_min);
}
}
private static void judge(Scanner in, char[] ch, int[] fr,int wpl_min) {
// TODO Auto-generated method stub
int n = ch.length;
int wpl=0;
int flag = 1;
String code [] = new String[n];
Jiedian root = new Jiedian();
char xi ;
for(int i=0;i<n;i++) {
xi = in.next().charAt(0);
code[i]= in.next();
int j = 0;
for(;j<n;j++) {
if (ch[j]==xi)
break;
}
if(j==n)
flag=0;
wpl =wpl+ code[i].length()*fr[j];
}
for(int i =0;i<n;i++) {
for(int j = 0;j<n;j++) {
if(i!=j&&code[i].startsWith(code[j]))
flag=0;
}
}
if (flag ==0||wpl!=wpl_min)
System.out.println("No");
else
System.out.println("Yes");
wpl=0;
}
private static int huffman_code_wpl( int[] fr) {
// TODO Auto-generated method stub
Arrays.sort(fr);
int n = fr.length;
Jiedian rt =null;
Jiedian [] jd = new Jiedian[n];
for(int i = 0; i<n ;i++) {
jd[i] = new Jiedian();
jd[i].num = fr[i];
}
for(int i = 0; i<n-1 ; i++) {
Jiedian root = new Jiedian();
root.left = jd[0];
root.right = jd[1];
root.num = jd[0].num+jd[1].num;
jd[1] = root;
jd[0] = new Jiedian();
jd[0].num = Integer.MAX_VALUE;
rt = root;
sort(jd);
}
int way = 0;
traversal(rt,way);
int wpl = sum;
sum=0;
return wpl;
}
private static void traversal(Jiedian rt,int way) {
// TODO Auto-generated method stub
if(rt==null);
else if(rt.left==null&&rt.right==null) {
sum=rt.num*way+sum;
}
else {
traversal(rt.left,way+1);
traversal(rt.right,way+1);
}
}
private static void sort(Jiedian[] jd) {
// TODO Auto-generated method stub
int n =jd.length;
Jiedian temp = new Jiedian();
for(int i = 0;i<n;i++) {
for(int j =i+1; j<n; j++) {
if(jd[j].num<jd[i].num) {
temp = jd[i];
jd[i] = jd[j];
jd[j] = temp;
}
}
}
}
}
class Jiedian{
Jiedian left =null;
Jiedian right = null;
String code =null;
int num = 0;
}