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  • Problem D

    Problem Description
    An entropy encoder is a data encoding method that achieves lossless data compression by encoding a message with “wasted” or “extra” information removed. In other words, entropy encoding removes information that was not necessary in the first place to accurately encode the message. A high degree of entropy implies a message with a great deal of wasted information; english text encoded in ASCII is an example of a message type that has very high entropy. Already compressed messages, such as JPEG graphics or ZIP archives, have very little entropy and do not benefit from further attempts at entropy encoding.

    English text encoded in ASCII has a high degree of entropy because all characters are encoded using the same number of bits, eight. It is a known fact that the letters E, L, N, R, S and T occur at a considerably higher frequency than do most other letters in english text. If a way could be found to encode just these letters with four bits, then the new encoding would be smaller, would contain all the original information, and would have less entropy. ASCII uses a fixed number of bits for a reason, however: it’s easy, since one is always dealing with a fixed number of bits to represent each possible glyph or character. How would an encoding scheme that used four bits for the above letters be able to distinguish between the four-bit codes and eight-bit codes? This seemingly difficult problem is solved using what is known as a “prefix-free variable-length” encoding.

    In such an encoding, any number of bits can be used to represent any glyph, and glyphs not present in the message are simply not encoded. However, in order to be able to recover the information, no bit pattern that encodes a glyph is allowed to be the prefix of any other encoding bit pattern. This allows the encoded bitstream to be read bit by bit, and whenever a set of bits is encountered that represents a glyph, that glyph can be decoded. If the prefix-free constraint was not enforced, then such a decoding would be impossible.

    Consider the text “AAAAABCD”. Using ASCII, encoding this would require 64 bits. If, instead, we encode “A” with the bit pattern “00”, “B” with “01”, “C” with “10”, and “D” with “11” then we can encode this text in only 16 bits; the resulting bit pattern would be “0000000000011011”. This is still a fixed-length encoding, however; we’re using two bits per glyph instead of eight. Since the glyph “A” occurs with greater frequency, could we do better by encoding it with fewer bits? In fact we can, but in order to maintain a prefix-free encoding, some of the other bit patterns will become longer than two bits. An optimal encoding is to encode “A” with “0”, “B” with “10”, “C” with “110”, and “D” with “111”. (This is clearly not the only optimal encoding, as it is obvious that the encodings for B, C and D could be interchanged freely for any given encoding without increasing the size of the final encoded message.) Using this encoding, the message encodes in only 13 bits to “0000010110111”, a compression ratio of 4.9 to 1 (that is, each bit in the final encoded message represents as much information as did 4.9 bits in the original encoding). Read through this bit pattern from left to right and you’ll see that the prefix-free encoding makes it simple to decode this into the original text even though the codes have varying bit lengths.

    As a second example, consider the text “THE CAT IN THE HAT”. In this text, the letter “T” and the space character both occur with the highest frequency, so they will clearly have the shortest encoding bit patterns in an optimal encoding. The letters “C”, “I’ and “N” only occur once, however, so they will have the longest codes.

    There are many possible sets of prefix-free variable-length bit patterns that would yield the optimal encoding, that is, that would allow the text to be encoded in the fewest number of bits. One such optimal encoding is to encode spaces with “00”, “A” with “100”, “C” with “1110”, “E” with “1111”, “H” with “110”, “I” with “1010”, “N” with “1011” and “T” with “01”. The optimal encoding therefore requires only 51 bits compared to the 144 that would be necessary to encode the message with 8-bit ASCII encoding, a compression ratio of 2.8 to 1.
     

    Input
    The input file will contain a list of text strings, one per line. The text strings will consist only of uppercase alphanumeric characters and underscores (which are used in place of spaces). The end of the input will be signalled by a line containing only the word “END” as the text string. This line should not be processed.
     

    Output
    For each text string in the input, output the length in bits of the 8-bit ASCII encoding, the length in bits of an optimal prefix-free variable-length encoding, and the compression ratio accurate to one decimal point.
     

    Sample Input
    AAAAABCD
    THE_CAT_IN_THE_HAT
    END
     

    Sample Output
    64 13 4.9
    144 51 2.8
    题意:给你一行字符,判断里面字符出现的次数,出现的次数作为该字符在哈夫曼数里的权值;
    哈夫曼树:就是一种压缩代码的东西;二叉树左儿子为0,右儿子为1,每个字符从根节点向上依次按照树枝的值来构建代码,所以每个数的代码都是唯一的;
    解题过程:看到这个题真是头痛,题目太长了,但是看到做的人不少而且正确率不低,所以做做试试,晚自习看了一晚上,看懂了大题题意,但是不知道咋写,第二天问了同学,说得用哈夫曼数,我去。。。。。。听都没听说过,没办法了,查了查哈曼树是什么,光是看代码怎么构建的就看了一天,下午有绊绊卡卡的写出来,还是有地方不明白,回头看了几眼原来的代码;
    感悟:自己懂得东西太少了,别人都已经会的东西,自己还需要查资料,但是我是不会放弃的,人一我百,人百我万!
    代码(构建的哈夫曼树多少有点水分^o^)

    #include
    #include
    #include
    #include
    #include
    #include
    #define maxn 20005
    using namespace std;
    typedef struct tree
    {
        int leaf;//记录字母出现次数
        int pa;//记录父亲结点
    };
    tree tr[80];//80个节点应该够了
    void huffman(int n)
    {
        int m1,m2,x1,x2,t,i,j;//m1,m2记录两个权值较小的结点,x1,x2记录对应下标
        for(i=0;i
        {
            m1=m2=0x3f3f3f3f;//初始化为无穷大;
            for(j=0;j
            {
                if(m1>tr[j].leaf&&tr[j].pa==-1&&tr[j].leaf!=0)//后两个条件是这个数
                //没有被用过,并且出现的次数比0大;
                {
                    m2=m1;
                    m1=tr[j].leaf;
                    x2=x1;
                    x1=j;
                }
                else if(m2>tr[j].leaf&&tr[j].pa==-1&&tr[j].leaf!=0)
                {
                    x2=j;
                    m2=tr[j].leaf;
                }
            }
            if(m2!=0x3f3f3f3f)//不为初始值就是找到了
            {
                t=n+i;
                tr[t].leaf=m1+m2;//这就是两个数上面的那个新的节点
                tr[x1].pa=tr[x2].pa=t;//删除两个以最小的节点
            }
        }
    }
    int main()
    {
        //freopen("in.txt", "r", stdin);
        int i,l,c,p,ans;
        char s[maxn];
        while(scanf("%s",s))
        {
            if(strcmp(s,"END")==0)//结束标志
              break;
            for(i=0;i<80;i++)//初始化
            {
                tr[i].leaf=0;
                tr[i].pa=-1;
            }
            l=strlen(s);//记录字符串长度
            for(i=0;i
                if(s[i]=='_')
                    tr[26].leaf++;//建立字符与数字下标的映射
                else
                    tr[s[i]-'A'].leaf++;
            huffman(27);//建立哈弗曼树
            for(i=0;i<=26;i++)
            {
                c=0;//c记录编码所需长度
                if(tr[i].leaf!=0)
                {
                    p=i;
                    while(tr[p].pa!=-1)
                    {
                        c++;
                        p=tr[p].pa;
                    }
                    if(c==0)//如果为根节点长度为1
                        c=1;
                    tr[i].leaf=c;//将data赋值为所需字节数。方便下面算总数
                }
            }
            ans=0;//ans记录所需总字节数
            for(i=0;i
            {
                if(s[i]=='_')
                    ans+=tr[26].leaf;
                else
                    ans+=tr[s[i]-'A'].leaf;
            }
            printf("%d %d %.1f ",8*l,ans,8.0*l*1.0/ans);//这个位置开始用的是%.1lf就WA了 应该是编译器的问题
        }
        return 0;
    }
    我每天都在努力,只是想证明我是认真的活着.
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  • 原文地址:https://www.cnblogs.com/wuwangchuxin0924/p/5781667.html
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