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  • 树9 Huffman Codes

    题目:https://pintia.cn/problem-sets/1268384564738605056/problems/1278908289143574530

    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 (2N63), 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


    题解:

    https://www.cnblogs.com/liangchao/p/4286598.html
    代码:


    #include <iostream>
    #include <string>
    #include <algorithm>    //使用sort函数
    #include <map>
    #include <queue>
    using namespace std;
    
    #define MinData 0
    #define MaxInputSize 2000
    
    //用PAIR来代替pair<char, string>
    typedef pair<char, string> PAIR;
    
    //哈弗曼树结构
    typedef struct HuffmanTreeNode{
        int weight;
        HuffmanTreeNode* leftweight;
        HuffmanTreeNode* rightweight;
    }*pHuffmanTree, nHuffmanTree;
    
    //最小堆结构
    typedef struct HeapStruct{
        pHuffmanTree Elememts;    //存储堆元素的数组
        int Size;    //堆当前元素个数
        int Capacity;    //堆得最大容量
    }*pMinHeap, nMinHeap;
    
    
    
    pMinHeap CreateMinHeap( int [], int );
    pMinHeap CreateEmptyMinHeap( int );
    void InsertMinHeap( pMinHeap, nHuffmanTree );
    pHuffmanTree DeleteMinHeap( pMinHeap);
    pHuffmanTree GetHuffmanRootWeight( pMinHeap );
    void GetWPLValue( pHuffmanTree pHT, int layer, int *wpl);
    
    //比较函数,按编码长度排序
    int cmp( const PAIR &x, const PAIR &y )
    {
        return x.second.size() < y.second.size();
    }
    
    int main()
    {
        int charNum;
        cin >> charNum;
        char *arChar = new char[ charNum ];
        int *arWeight = new int[ charNum ];
        int i;
        for( i = 0; i < charNum; i++ )
        {
            cin >> arChar[i] >> arWeight[i];
        }
        pMinHeap minH = CreateMinHeap( arWeight, charNum );
        pHuffmanTree pHT = GetHuffmanRootWeight( minH );
        int originWPL = 0;
        GetWPLValue( pHT, 0, &originWPL );
        int stuNum;
        cin >> stuNum;
        int j;
        char temp;
        string *str = new string[charNum];
        int stuWPL;
        string outputstr = "";
        for ( i = 0; i < stuNum; i++ )
        {
            stuWPL = 0;
            //vector定义
            vector<PAIR> checkVec;
            for ( j = 0; j < charNum; j++ )
            {
                cin >> temp >> str[j];
                //向vector中添加元素
                checkVec.push_back( make_pair( temp, str[j] ) );
                stuWPL += arWeight[j] * str[j].length();
            }
            //按编码长度排序
            sort( checkVec.begin(), checkVec.end(), cmp );
            int cmp1, cmp2;
            if ( stuWPL == originWPL )
            {
                bool flag = true;
                for( cmp1 = 0; cmp1 < charNum; cmp1++ )
                {
                    string tmpstr = checkVec[cmp1].second;
                    for ( cmp2 = cmp1 + 1; cmp2 < charNum; cmp2++ )
                    {
                        if ( checkVec[cmp2].second.substr( 0, tmpstr.size() ) == tmpstr )
                        {
                            flag = false;
                        }
                    }
                }
                if ( flag == true )
                {
                    cout << "Yes" << endl;
                }
                else
                {
                    cout << "No" << endl;
                }
            }
            else
            {
                cout << "No" << endl;
            }
        }
        return 0;
    }
    
    pMinHeap CreateMinHeap( int weight[], int len )
    {
        int i;
        pMinHeap minH = CreateEmptyMinHeap( MaxInputSize );
        nHuffmanTree nHT;
        for ( i = 0; i < len; i++ )    //顺序插入构造最小堆(更有效率的方式为先建立完全二叉树,再调整为最小堆)
        {
            nHT.weight = weight[i];
            nHT.leftweight = NULL;
            nHT.rightweight = NULL;
            InsertMinHeap( minH, nHT );
        }
        return minH;
    }
    pMinHeap CreateEmptyMinHeap( int MaxSize )
    {
        pMinHeap minH = ( pMinHeap )malloc( sizeof( nMinHeap ) );
        minH->Elememts = ( pHuffmanTree )malloc( ( MaxSize + 1 ) * sizeof( nHuffmanTree ) );
        minH->Size = 0;
        minH->Capacity = MaxInputSize;
        minH->Elememts[0].weight = MinData;    //哨兵元素
        return minH;
    }
    
    void InsertMinHeap( pMinHeap minH, nHuffmanTree nHT )
    {
        //将元素item插入最小堆,其中minH->Elements[0]定义为哨兵元素
        int i;
        if ( minH->Size >= minH->Capacity )
        {
            cout << "Heap Is Full!" << endl;
            return;
        }
        i = ++minH->Size;    //i指向插入后堆中的最后一个元素的位置(该结点此时为空结点)
        for( ; minH->Elememts[i/2].weight > nHT.weight; i/=2 )
        {
            minH->Elememts[i] = minH->Elememts[i/2];
        }
        minH->Elememts[i] = nHT;
    }
    
    pHuffmanTree DeleteMinHeap( pMinHeap minH)
    {
        //从最小堆H中取出键值为最小的元素,并删除一个结点
        int parentNode, childNode;
        nHuffmanTree temp;
        pHuffmanTree minNode;
        if ( minH->Size >= minH->Capacity )
        {
            cout << "Heap Is Full!" << endl;
        }
        minNode = ( pHuffmanTree )malloc( sizeof( nHuffmanTree ) );
        *minNode = minH->Elememts[1];    //取出根结点的最小值
        temp = minH->Elememts[ minH->Size-- ];  //用最小堆的最后一个元素从根结点开始向上过滤下层结点
        for ( parentNode = 1; parentNode * 2 <= minH->Size; parentNode = childNode )
        {
            //找出当前parentNode结点的最小子结点
            childNode = parentNode * 2;
            if ( childNode != minH->Size && minH->Elememts[childNode].weight > minH->Elememts[childNode + 1].weight  )    //当存在右子结点,且右子节点小于左子节点时
            {
                childNode++;
            }
            if ( temp.weight <= minH->Elememts[childNode].weight )
            {
                break;
            }
            else
            {
                minH->Elememts[parentNode] = minH->Elememts[childNode];
            }
        }
        minH->Elememts[ parentNode ] = temp;
        return minNode;
    }
    
    pHuffmanTree GetHuffmanRootWeight( pMinHeap minH)
    {
        int i;
        pHuffmanTree pHT;
        int times = minH->Size;
        for ( i = 1; i < times; i++ )    //执行初始Size-1次合并
        {
            pHT = ( pHuffmanTree )malloc( sizeof( nHuffmanTree ) );    //使用这种方法建树会出现malloc多余内存的情况
            pHT->leftweight = DeleteMinHeap( minH );    //Delete操作将使得ElementSize = ElementSize - 1
            pHT->rightweight = DeleteMinHeap( minH );
            pHT->weight = pHT->leftweight->weight + pHT->rightweight->weight;
            InsertMinHeap( minH, *pHT );
        }
        pHT =  DeleteMinHeap( minH );
        return pHT;
    }
    
    void GetWPLValue( pHuffmanTree pHT, int layer, int *wpl)
    {
        if ( pHT->leftweight == NULL && pHT->rightweight == NULL )
        {
            (*wpl) += layer * pHT->weight;
        }
        else    //非叶结点必有两个子结点
        {
            GetWPLValue( pHT->leftweight, layer + 1, wpl );
            GetWPLValue( pHT->rightweight, layer + 1, wpl );
        }
    }
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  • 原文地址:https://www.cnblogs.com/simon-chou/p/13619953.html
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