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  • Move-to-front(MTF) and Run-lenght encoding(RLE) algorithms

      mtf算法(关于该算法:https://www2.cs.duke.edu/csed/algoprobs/beta/bw1.html):

    #include <stdio.h>
    #include <string.h>
    #include <stdlib.h>
    
    void mtf_encode(const char *s, unsigned len, int *code)
    {
        int pos = 0, num, trace = 0;
        unsigned i, j, k;
        int is_hav[26], stack[26];
    	
        memset(is_hav, 0, 26);
    	
        while (*s)
        {
            if (!is_hav[*s - 'a'])
            {
                stack[pos] = *s - 'a' + 1;
                is_hav[*s - 'a'] = 1;
                ++pos;
                code[trace++] = pos;
            }
            else
            {
                j = 0;
                while (stack[j] != *s - 'a' + 1) j++;
                code[trace++] = j + 1, num = stack[j];
                while (j > 0) stack[j] = stack[j - 1], j--;
                stack[j] = num;
            }
            s++;
        }
    }
    
    void print_mtf_encode(int*code, int len)
    {
        unsigned i;
        printf("[");
        for (i = 0; i < len; i++)
        {
            printf("%d", code[i]);
            if (i != len - 1)
                printf(", ");
        }
        printf("]
    ");
    }
    
    int main()
    {
        char s[] = "abcabcaaaaaaaab";
        int len = strlen(s);
        int *code = (int *)malloc(sizeof(int) * len);
        mtf_encode(s, len, code);
        print_mtf_encode(code, len);
        free(code);
        return 0;
    }

      上面算法不是错的,根据题意和不同的思考角度可以编写不同的mtf算法,按字典序排序后的mtf算法(只统计出现的字符):

    void mtf_encode(const char *s, unsigned len, int *code)
    {
        int num, trace = 0;
        unsigned i, pos = 0;
        int is_hav[26], stack[26];
    	
        memset(is_hav, 0, 26);
    	
        for (i = 0; i < len; i++)
    	if (!is_hav[s[i] - 'a'])
                is_hav[s[i] - 'a'] = s[i] - 'a' + 1;
    	
        for (i = 0; i < 26; i++)
    	if (is_hav[i] != 0)
    	    stack[pos++] = is_hav[i];
    
        while (*s)
        {
            pos = 0;
            while (stack[pos] != *s++ - 'a' + 1) pos++;
            code[trace++] = pos + 1, num = stack[j];
            while (pos > 0) stack[pos] = stack[pos - 1], pos--;
            stack[pos] = num;
        }
    }

      rle算法(了解该算法:https://en.wikipedia.org/wiki/Run-length_encoding):

    const char*digits = "0123456789";
    
    void comp_bit(char *rle, int *i, int num)
    {
        if (num > 9)
            comp_bit(rle, i, num / 10);
        rle[(*i)++] = digits[num%10];
    }
    
    void rle_encode(const char *s, char *rle)
    {
        unsigned i = 0, j = 1;
        rle[i++] = *s++;
        while (*s)
        {
            if (rle[i - 1] != *s)
            {
                comp_bit(rle, &i, j);
                rle[i++] = *s, j = 1;
            }
            else j++;
            s++;
        }
        comp_bit(rle, &i, j);
    }
    

      

      

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  • 原文地址:https://www.cnblogs.com/darkchii/p/10060029.html
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