05-树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
Sample Output:
Yes
Yes
No
No
题目
输入的第一行代表数的个数
第二行为给定的元素和权值
第三行为要判定的组数
后面接着每n行一个组
解题关键
1.建立最小堆，把第二行的数据逐个插入最小堆，再另取一个列表保留权值；
2.利用霍夫曼树的建立方法，把最小堆里的头元素两两组合再插回，执行n-1次，便将最小堆转化为霍夫曼树；
3.通过此霍夫曼树，求出最小权值
两部将题目给定的元素建立成了霍夫曼树
接下来要判断后面给的数是否为霍夫曼树
1.最小权值相同
字符串长度*给定字母的权值大小就是该路径的权值；
全部相加即为总权值，与已求权值比较即可；
2.不会出现前缀码重复
构建一个二叉树，若有值则标记；
后面路径路过此值即为前缀码重复；
后面路径终点不为叶节点，也为前缀码重复；
#include<iostream>
#include<cstring>
using namespace std;
#define MaxNum 64
struct TreeNode{ //树
int Weight =0; //初始化为0
TreeNode *Left = nullptr;
TreeNode *Right = nullptr; //左右为指向下一个tree对象的空指针
};
struct HeapNode{// 堆结构
TreeNode Data[MaxNum]; //data内放着树结构
int Size =0; //初始大小为0
};
HeapNode * CreateHeap(){ //建树——返回指向树根节点的指针
HeapNode *H = new(HeapNode);
H->Data[0].Weight = -1; //哨兵的weight设置为-1
return H; //建堆——一次即可
}
TreeNode *DeleteMin(HeapNode *H){ //返回堆定的指针
int Parent = 0 , Child=0;
TreeNode temp;
TreeNode *MinItem = new(TreeNode); //新建一个minitem
*MinItem = H->Data[1]; //保存堆顶
temp = H->Data[(H->Size)--]; //拿到堆底的对象，并且size-1
for(Parent=1;Parent *2 <= H->Size; Parent=Child)
{
Child= Parent*2;
//Size就是为了作为边界
if(Child!=H->Size && (H->Data[Child].Weight>H->Data[Child+1].Weight))
Child++;
if(temp.Weight<=H->Data[Child].Weight) break;
else H->Data[Parent]=H->Data[Child]; //下沉
}
//break出来说明找到合适位置
H->Data[Parent] = temp; //进行赋值
return MinItem; //返回堆定成员指针
}
//插入
void Insert(HeapNode *H, TreeNode *item){ //将一个树成员插入堆中
int i=0;
i= ++(H->Size); //i为当前总容量
//item为指向树成员的指针
for (;item ->Weight<H->Data[i/2].Weight;i/=2) //插入最后，然后上浮
//当要插入的成员小于父节点，上浮
{
H->Data[i]=H->Data[i/2];
}
H->Data[i] = *item; //将item所指成员放入堆中
}
//
//数组存放value
//因为后面是按顺序给的
//所以顺序保存每个字母的权值，后面遍历数组就可以得到权值
//计算总权值比较方便
HeapNode * ReadData(int N, HeapNode *H, int A[])
{ //读取数据，存放至A，并且插入树中，循环N次
char s;
int value;
for(int i=0;i<N;++i)
{
cin>>s;
cin>>value;
A[i]=value;
TreeNode* T = new(TreeNode); //新建动态内存，用到指针
T->Weight = value;
Insert(H,T); //insert指针更方便
}
return H; //返回堆头
}
TreeNode *Huffman(HeapNode *H){ //把堆变为哈夫曼树
TreeNode *T = nullptr; //初始空指针
int num = H->Size; //一共有num个元素
//进行n-1次合并
for(int i=0;i<num-1;i++)
{
T= new(TreeNode);
T->Left = DeleteMin(H);
T->Right = DeleteMin(H);
T->Weight=T->Left->Weight + T->Right->Weight;
Insert(H,T); //T为指针——insert(H,T)
//要用动态内存创造空间时——用指针
}
T= DeleteMin(H);
return T; //返回指向堆头的指针
//
}
//计算树H的权值——n从0开始，遇到叶节点，返回值，遇到根节点，把左右加起来返回
int WPL(TreeNode *H, int n)
{
if(!H ->Right) return H->Weight * n;
else return WPL(H->Left,n+1) +WPL(H->Right,n+1);
}
struct JNode{
// 每个点非0即1 flag为1代表这里已经有一个前辈
//住进去了只要我完全走了前辈走的路，就重复出现
int flag=0; //0表示不为叶节点，1表示为叶节点
JNode *Left=nullptr;
JNode *Right =nullptr;
};
//判断编码是否符合前缀码要求
bool Judge(string s, JNode *J)
{
int i=0;
for (;i <s.length();++i) //堆给定string的长度循环
{
if(s[i]=='0'){ //遇到0
if(J->Left==nullptr){ //没有路
JNode *j = new(JNode); //创建一个树结点
J->Left =j; //0为把新建的树挂到左边
}
else{ //如果有路
//这条路为叶节点
if(J->Left->flag==1) return false; //则返回错误
//不为叶结点则什么都不做，继续走
}
J=J->Left; //走到左边的点
}
else { //遇到1
if(J->Right==nullptr){
JNode *j=new(JNode); //这个用法
J->Right = j;
}
else {
if(J->Right ->flag==1) return false;
}
J=J->Right;
}
}
//循环结束
J->flag =1; //将找到的节点设为1；
if(J->Left==nullptr&&J->Right==nullptr){ //如果该点确实为叶节点——没有左右子树
return true;
}
else return false;
}
int main(){
int N=0, n=0;
cin>>N;
HeapNode *H = CreateHeap(); //H指向一个新建堆
int Value[MaxNum] = {};
ReadData(N,H,Value); //把N个数， 插入堆H， 保存至数组Value
TreeNode *T = Huffman(H); //把堆H按霍夫曼组合，得到树头，给T
int CodeLen=WPL(T,0); //从树头开始，计算权值
cin>>n; //读取要判断的组数个数
string temp;
char c;
bool result;
for(int i=0;i<n;++i) //n组
{
int count=0;
int Flag=0;
JNode *J = new(JNode); //新建一个树头
//这里写错，真的离谱
for(int j=0;j<N;++j){ //第j个数的权值放在value[j]
cin>>c>>temp; //读入字符，字符串
count+=temp.length()*Value[j];
//Value存放 权值，temp的长度*权值就为该字母的权值
if(!Flag){
result = Judge(temp, J); //判断
if(!result){ //出现一行不符合前缀码要求
Flag=1; //标记无需计算，读完即可
}
}
}
delete J; //释放J的内存
if(result && count == CodeLen){ //权值相等,并且经过上面前缀码的判断
cout<<"Yes"<<endl;
}
else{
cout<<"No"<<endl;
}
}
return 0;
}
