陈越《数据结构》第七讲 图中

Tree Traversals Again

1086.Tree Traversals Again (25)
An inorder binary tree traversal can be implemented in a non-recursive way with a stack. For example,suppose that when a 6-node binary tree (with the keys numbered from 1 to 6) is traversed,the stack operations are: push(1); push(2); push(3); pop(); pop(); push(4); pop(); pop(); push(5); push(6); pop(); pop(). Then a unique binary tree (shown in Figure 1) can be generated from this sequence of operations. Your task is to give the postorder traversal sequence of this tree.

Input Specification:

Each input file contains one test case. For each case,the first line contains a positive integer N (<=30) which is the total number of nodes in a tree (and hence the nodes are numbered from 1 to N). Then 2N lines follow,each describes a stack operation in the format: “Push X” where X is the index of the node being pushed onto the stack; or “Pop” meaning to pop one node from the stack.

Output Specification:

For each test case,print the postorder traversal sequence of the corresponding tree in one line. A solution is guaranteed to exist. All the numbers must be separated by exactly one space,and there must be no extra space at the end of the line.

Sample Input:
6
Push 1
Push 2
Push 3
Pop
Pop
Push 4
Pop
Pop
Push 5
Push 6
Pop
Pop
Sample Output:
3 4 2 6 5 1

题目大意：
用栈的形式给出一棵二叉树的建立的顺序，求这棵二叉树的后序遍历
分析：栈实现的是二叉树的中序遍历（左根右），而每次push入值的顺序是二叉树的前序遍历（根左右），所以该题可以用二叉树前序和中序转后序的方法做~~

#include <cstdio>
#include <stack>
using namespace std;

int preorder[35],inorder[35];
int n,preid = 0,inid = 0,cnt = 0;
int get(){
    char s[10];
    scanf("%s",s);
    if (s[1] == 'o') return -1;
    int a;
    scanf("%d",&a);
    return a;
}
void build(int preb,int pree,int inb,int ine){
    if (preb > pree) return;
    int root = preorder[preb];
    int inroot = inb;
    while (inorder[inroot] != root) ++inroot;
    build(preb+1,preb+inroot-inb,inb,inroot-1);
    build(preb+inroot-inb+1,pree,inroot+1,ine);
    if (cnt++ != 0) putchar(' ');
    printf("%d",root);
}
int main(){
    scanf("%d",&n);
    stack<int> st;
    for (int i = 0; i < n*2; ++i){
        int a = get();
        if (a != -1){
            st.push(a);
            preorder[preid++] = a;
        }else{
            inorder[inid++] = st.top();
            st.pop();
        }
    }
    build(0,n-1,0,n-1);
    return 0;
}

Complete Binary Search Tree

A Binary Search Tree (BST) is recursively defined as a binary tree which has the following properties:

The left subtree of a node contains only nodes with keys less than the node’s key.
The right subtree of a node contains only nodes with keys greater than or equal to the node’s key.
Both the left and right subtrees must also be binary search trees.
A Complete Binary Tree (CBT) is a tree that is completely filled,with the possible exception of the bottom level,which is filled from left to right.

Now given a sequence of distinct non-negative integer keys,a unique BST can be constructed if it is required that the tree must also be a CBT. You are supposed to output the level order traversal sequence of this BST.

Input Specification:

Each input file contains one test case. For each case,the first line contains a positive integer N (<=1000). Then N distinct non-negative integer keys are given in the next line. All the numbers in a line are separated by a space and are no greater than 2000.

Output Specification:

For each test case,print in one line the level order traversal sequence of the corresponding complete binary search tree. All the numbers in a line must be separated by a space,and there must be no extra space at the end of the line.

Sample Input:
10
1 2 3 4 5 6 7 8 9 0
Sample Output:
6 3 8 1 5 7 9 0 2 4

题意：
给一串构成树的序列，已知该树是完全二叉搜索树，求它的层序遍历的序列
分析：
总得概括来说，已知中序，可求root下标，可以求出层序。
1. 因为二叉搜索树的中序满足：是一组序列的从小到大排列，所以只需排序所给序列即可得到中序
2. 因为根据完全二叉树的结点数，可以求出它的根结点在中序中对应的下标
3. 如此，已知了中序，又可以根据结点数求出根结点的下标，就可以递归求出左右子树的根结点的下标
4. i结点的左孩子为2 * i + 1，右孩子2 * i + 2，就可以根据结点下标和中序数组赋值level数组
5. 最后输出所有结点的层序数组level

图1

图2

图3

#include <cstdio>
#include <cstdlib>
#include <math.h>

const int MaxNum = 1005;
int node[MaxNum];
int tree[MaxNum];

int cmp(const void *a,const void *b)
{/*以图2为基准*/
    int *pa = (int *)a;
    int *pb = (int *)b;
    return *pa-*pb;
}
int GetLeftLength(int n)
{/*以图3为基准*/
    int h,x,l;
    /*转换成以2为底*/
    h = log((double)(n + 1))/log (2.0);
    x = n + 1 - pow(2.0,h);
    if (x > pow(2.0,h - 1)) {
        x = pow(2.0,h - 1);
    }
    l = pow(2.0,h - 1) - 1 + x;
    return l;
}

void solve(int ALeft,int ARight,int TRoot)
{/*初始调用为solve(0,N-1,0)*/
/*以图1为基准*/
    int n,L,LeftTRoot,RightTRoot;
    n = ARight - ALeft + 1;
    if(n==0) return;        /*考虑完美二叉树的情况*/
    L = GetLeftLength(n);    /*计算出n个节点的树，其左子树有多少个节点*/
    tree[TRoot] = node[ALeft + L];
    LeftTRoot = TRoot * 2 + 1;
    RightTRoot = LeftTRoot + 1;
    solve(ALeft,ALeft + L -1,LeftTRoot);
    solve(ALeft + L + 1,ARight,RightTRoot);

}

int main()
{
    int i,n;
    /*输入*/
    scanf("%d",&n);
    for(i=0;i<n;++i){
        scanf("%d",&node[i]);
    }
    /*从大到小排序*/
    qsort(node,n,sizeof(int),cmp);
    /*排序成完全二叉树*/
    solve(0,0);
    /*输出*/
    printf("%d",tree[0]);
    for(i=1;i<n;++i){
        printf(" %d",tree[i]);
    }
    printf("\n");
    system("pause");
    return 0;
}

Huffman Codes

PTA - Data Structures and Algorithms (English) - 5-9

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 ${'a'=0,'u'=011,'z'=001}$ 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 $\color{red}{not}$ $\color{red}{necessarily}$ $\color{red}{generated}$ $\color{red}{by}$ $\color{red}{Huffman}$ $\color{red}{algorithm}$. Any prefix code with code length being optimal is considered correct.

Sample Input:
7 //结点数目num
A 1 B 1 C 1 D 3 E 3 F 6 G 6 //每个结点数据data及出现的次数weight
4 //测试数据的组数checkNum
A 00000 //之后的 4*7行 是结点数据ch及其编码s
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

题目分析：

这是一道考察“哈夫曼编码”的问题，但是这里不一定非要把哈夫曼树构造出来。Note: The optimal solution is not necessarily generated by Huffman algorithm
- 输入：第一行是结点数目num；第二行是每个结点数据data及出现的次数weight；第三行是测试数据的组数checkNum；第四行及以后是结点数据ch及编码s。

• 输出：对于每一组测试数据，输出编码是否符合“哈夫曼编码”，是则输出Yes，否则输出No。

• 符合“哈夫曼编码”需要符合两个条件：
  ①$\color{red}{WPL最小}$
  ②$\color{red}{编码的前缀不能是其他编码的前缀。}$

例1：最优编码不一定通过Huffman算法得到。给定4个字符及其出现频率：$A:1; B:1; C:2; D:2$。下面哪一套不是用Huffman算法得到的正确的编码？
A. A:000; B:001; C:01; D:1
B. A:10; B:11; C:00; D:01
C. A:00; B:10; C:01; D:11
D. A:111; B:001; C:10; D:1
解析：C
根据答案，画出每个答案的Huffman tree，再一个个的对答案是否正确。

例2：Huffman Codes的特点：

1. 最优编码 —— 总长度（WPL）最小；
2. 无歧义解码 —— 前缀码：数据仅存于叶子结点；
3. 没有度为1 的结点 —— $\color{red}{满足1 、2 则必然有3}$。
   注意：$\color{red}{满足2 、3 可不一定有1 }$！比如：
   

解法：

解法转自：http://www.cnblogs.com/clevercong/p/4193370.html
1) map 用于存放：A 1 B 1 C 1 D 3 E 3 F 6 G 6 //每个结点的数据data及出现的次数(权值)weight

2) 使用C++标准库中的优先队列：priority_queue，引入头文件 #include 。优先队列底层由堆实现，数据放入队列后，会自动按照“优先级”排好顺序。

#include <map>
#include <queue>

map<char,int> myMap;
priority_queue<int,vector<int>,greater<int> >pq;  //小顶堆

for(int i=0; i<num; i++)  // 输入结点的数据c[i]、权值f[i]
{
    cin >> c[i] >> f[i];
    myMap[c[i]] = f[i];  // 映射
    pq.push(f[i]);  // 向队列中添加元素
}

3） 计算$WPL$的值，从priority_queue中取出两个元素，相加之后再放回队列里。

// 计算WPL的值
int myWpl = 0;
while(!pq.empty())
{
    int myTop = pq.top();
    pq.pop();
    if(!pq.empty())
    {
        int myTop2 = pq.top();
        pq.pop();
        pq.push(myTop + myTop2);
        int m = myTop + myTop2;
        myWpl += m;  //每次加m(子节点权值重复加入) 等效于 路径长度*权值
    }
}

4) 测试数据需按编码排序，但标准库并没有为map制定sort函数，因此我们用vector装载pair类型，既可以模仿出map的功能，又可以用vector的排序函数。

#include <algorithm> // sort()
typedef pair<char,string> PAIR;  // 用PAIR来代替pair<char,string> (编码类型:string)

// cmp()：自定义按什么内容或大小顺序排序
// 这里是按编码的长度排序
int cmp(const PAIR& x,const PAIR& y)
{
    return x.second.size() < y.second.size();
}
// vector + pair<,> 模仿 map
vector<PAIR> checkVec;
checkVec.push_back(make_pair(ch,s));  // 向vector中添加元素
sort(checkVec.begin(),checkVec.end(),cmp);  // 按照编码的长度排序

5) 判断前缀问题：substr函数，取字符串中的一段并与当前编码进行比较。

bool flag = true;  //已符合条件一：wpl最小
for(int i=0; i<num; i++)
{
    string tmp = checkVec[i].second;
　　for(int j=i+1; j<num; j++)
    {
        if(checkVec[j].second.substr(0,tmp.size())==tmp)
            flag = false;
    }
}

完整代码：

#include <iostream>
#include <algorithm> // 排序函数 sort()
#include <map>
#include <queue>
using namespace std;

typedef pair<char,string> PAIR;  // + vector来模仿 map

int cmp(const PAIR& x,const PAIR& y)  // 自定义让sort()按哪种方式排序
{
    return x.second.size() < y.second.size();
}

int main()
{
    int num;
    cin >> num;
    char *c = new char[num];
    int *f = new int[num];
    map<char,int> myMap;  // 用来存节点数据及权值，并构成映射
    // 使用优级队列
    priority_queue<int,greater<int> >pq;

    for(int i=0; i<num; i++)  // 输入结点及出现次数(权值)
    {
        cin >> c[i] >> f[i];
        myMap[c[i]] = f[i];
        pq.push(f[i]);  // 将权值压入优先队列
    }
    // 计算WPL的值
    int myWpl = 0;
    while(!pq.empty())
    {
        int myTop = pq.top();
        pq.pop();
        if(!pq.empty())
        {
            int myTop2 = pq.top();
            pq.pop();
            pq.push(myTop + myTop2);
            int m = myTop + myTop2;
            myWpl += m;
        }
    }
    // 输入测试数据
    int checkNum;  // 测试数据的组数
    cin >> checkNum;
    for(int i=0; i<checkNum; i++)
    {
        int wpl = 0;
        char ch;
        string s;
        // vector + PAIR 模仿 map，使其可排序
        vector<PAIR> checkVec;
        for(int j=0; j<num; j++)
        {
            cin >> ch >> s;
            checkVec.push_back(make_pair(ch,s));  // 向vector中添加测试数据及其编码
            wpl += s.size() * myMap[ch];
        }
        sort(checkVec.begin(),cmp);  // 按照编码长度排序
        if(wpl != myWpl)
        {
            cout << "No" << endl;
            continue;
        }
        else
        {
            bool flag = true;  // 表示已满足条件一：wpl最小(wpl==myWpl)

            //条件二：编码的前缀不能是其他编码的前缀：substr()
            for(int i=0; i<num; i++)
            {
                string tmp = checkVec[i].second;
                for(int j=i+1; j<num; j++)
                {
                    if(checkVec[j].second.substr(0,tmp.size())==tmp)
                        flag = false;
                }
            }
            if(flag == true)
                cout << "Yes" << endl;
            else
                cout << "No" << endl;
            continue;
        }
        cout << "Yes" << endl;
    }
    return 0;
}

总结的程序

#include <iostream>
#include <string>
using namespace std;
struct Heap
{
    int *data;
    int size = 0;
};
struct cnode
{
    int tag = 0;
    cnode *left = nullptr;
    cnode *right = nullptr;
};

Heap *Init_Heap(int n); //初始化小根堆（利用静态链表的逻辑结构）；
void Insert_Heap(Heap *H,int x);   //把元素依次插入小根堆；
int Delmin_Heap(Heap *H);   //删除小根堆中的最小元素；
int Cal_Wpl(Heap *H);   //计算huffman编码的wpl；
int wpl_prefix(int *sample,int n,int &re);    //计算待测编码的wpl及判断是否为前缀编码；

/*思路：要判断是否，需要解决两个问题： 1）编码wpl等于huffman编码的wpl； 2)待测编码是前缀编码。 问题1： 首先要求出标准wpl。观察huffman树，我们发现其wpl是非叶子结点权值和。 于是，我们无需构造出huffman树来求权值(麻烦点），通过模拟树的构造过程， 理由小根堆的特点求出wpl； 而待测编码的wpl就是利用每个编码的字符串长度，乘以每个字符的权值得到； 问题2： 思路是构造一个二叉树出来，模拟huffman树。 1.让每个编码遍历树，结束时在当前结点设置标记为1； 2.如果遍历时，未结束时碰到了标记为1的结点，则不是前缀编码； 3.结束时，未碰到标记点（包括最后一个），是前缀编码。 */
int main()
{
    int n,i,val;
    char ch;
    cin >> n;
    int *sample = new int[n];   //每个字符权值存储在这里；
    Heap *H = Init_Heap(n); //初始化小根堆（利用静态链表的逻辑结构）；
    for (i = 0; i < n; i++)
    {
        cin >> ch >> val;
        sample[i] = val;
        Insert_Heap(H,val);    //把元素依次插入小根堆；
    }
    int best_wpl = Cal_Wpl(H);  //计算huffman编码的wpl；

    int m,wpl,re; //re是用来判断是否前缀编码，是为1，否为0；
    cin >> m;
    for (i = 0; i < m; i++)
    {
        wpl = wpl_prefix(sample,re);    //计算待测编码的wpl及判断是否为前缀编码；
        if (wpl == best_wpl && re == 1)
            cout << "Yes" << endl;
        else
            cout << "No" << endl;
    }
    delete sample;

    return 0;
}

Heap *Init_Heap(int n)
{
    Heap *H = new Heap;
    H->data = new int[n + 1];   //堆的下标从1开始（为了操作方便）；
    H->data[0] = -1;    //小根堆赋值最小值；

    return H;
}

void Insert_Heap(Heap *H,int x)    //把元素依次插入小根堆；
{
    int i = ++(H->size);
    for (; H->data[i / 2] > x; i /= 2)  //从下往上过滤；
        H->data[i] = H->data[i / 2];
    H->data[i] = x;

    return;
}

int Delmin_Heap(Heap *H)    //删除小根堆中的最小元素；
{
    int t = H->data[H->size--];
    int min = H->data[1];
    int pa,chi;
    for (pa = 1; pa * 2 <= H->size; pa = chi)   //从上往下过滤最小元素；
    {
        chi = pa * 2;
        if (chi != H->size && H->data[chi + 1] < H->data[chi])  //找到子结点的最小下标；
            chi++;
        if (t < H->data[chi])
            break;
        else
            H->data[pa] = H->data[chi];
    }
    H->data[pa] = t;    //赋值最小元素

    return min;
}

int Cal_Wpl(Heap *H)    //计算huffman编码的wpl；
{
    int sum = 0;
    if (H->size > 1)
    {
        int i,t1,t2,t;
        int m = H->size;
        for (i = 1; i < m; i++)
        {
            t1 = Delmin_Heap(H);    //模拟构造huffman树的过程，先取出两个最小值，相加，把和插入堆中；
            t2 = Delmin_Heap(H);
            t = t1 + t2;
            Insert_Heap(H,t);
            sum += t;
        }
    }
    else
        sum = H->data[0];

    return sum;
}

int wpl_prefix(int *sample,int &re) //计算待测编码的wpl及判断是否为前缀编码；
{
    int i,j,len,wpl = 0;
    char ch;
    string st;
    cnode *phead = new cnode;
    cnode *p = phead;
    re = 1;
    for (i = 0; i < n; i++)
    {
        cin >> ch >> st;    //此处计算wpl；
        len = st.length();
        wpl += len*sample[i];

        if (re != 0)    //此处判断是否前缀编码；
        {
            for (j = 0; j < len; j++)
            {
                if (st[j] == '0')   //0的话判断左边；
                {
                    if (p->left == nullptr) //左边空，新建结点；
                    {
                        cnode *q = new cnode;
                        p->left = q;
                    }
                    else
                    {
                        if (p->tag == 1)
                        {
                            re = 0;
                            break;
                        }
                    }
                    p = p->left;
                }
                else if (st[j] == '1')  //判断右边；
                {
                    if (p->right == nullptr)
                    {
                        cnode *q = new cnode;
                        p->right = q;
                    }
                    else
                    {
                        if (p->tag == 1)
                        {
                            re = 0;
                            break;
                        }
                    }
                    p = p->right;
                }
            }

            if (p->left == nullptr && p->right == nullptr)
                re = 1;
            else
                re = 0;
            if (p->tag == 0)    //注意此处需要判断，最后结点标记不为1才能赋值（待测编码有可能有相同的）；
                p->tag = 1;
            else
                re = 0;

            p = phead;
        }
    }

    return wpl;
}