Immediate Decodability

最新推荐文章于 2022-01-10 05:11:12 发布

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最新推荐文章于 2022-01-10 05:11:12 发布

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原文链接：http://www.cnblogs.com/NYNU-ACM/p/4237326.html

本文探讨了二进制编码的立即可解性概念，即一个符号的编码不能是另一个符号编码的前缀。介绍了如何通过程序判断一组二进制码是否满足立即可解性条件，并提供了具体的输入输出示例及代码实现。

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Description

An encoding of a set of symbols is said to be immediately decodable if no code for one symbol is the prefix of a code for another symbol. We will assume for this problem that all codes are in binary, that no two codes within a set of codes are the same, that each code has at least one bit and no more than ten bits, and that each set has at least two codes and no more than eight.

Examples: Assume an alphabet that has symbols {A, B, C, D}

The following code is immediately decodable:

A:01 B:10 C:0010 D:0000

but this one is not:

A:01 B:10 C:010 D:0000 (Note that A is a prefix of C)

Input

Write a program that accepts as input a series of groups of records from a data file. Each record in a group contains a collection of zeroes and ones representing a binary code for a different symbol. Each group is followed by a single separator record containing a single 9; the separator records are not part of the group. Each group is independent of other groups; the codes in one group are not related to codes in any other group (that is, each group is to be processed independently).

Output

For each group, your program should determine whether the codes in that group are immediately decodable, and should print a single output line giving the group number and stating whether the group is, or is not, immediately decodable.

The Sample Input describes the examples above.

Sample Input

Sample Output

Set 1 is immediately decodable
Set 2 is not immediately decodable

HINT

//（如果任意一个串，均不是其它串的前缀则表明，is immediately decodable，否则is not immediately decodable）

#include<stdio.h>
#include<string.h>
int main()
{
	char a[100][100];
	int t=0,i,j,t1,t2,k,s=1;
	while(gets(a[t]))
	{
		t++;
		while(gets(a[t]))
		{
			if(a[t][0]=='9')
				break;
			t++;
		}
		for(i=0;i<t;i++)
		{
			t1=strlen(a[i]);
			for(j=i+1;j<t;j++)
			{
				//t2=strlen(a[j]);
				for(k=0;k<t1;k++)
				{
					if(a[i][k]!=a[j][k])//(is immediately decodable)
						break;
				}
				if(k==t1)//(is not immediately decodable)
					break;
			}
			if(j!=t)//(is not immediately decodable)
				break;
		}
		if(i==t)       
            printf("Set %d is immediately decodable\n",s);
		else
            printf("Set %d is not immediately decodable\n",s);
	
		//for(i=0;i<t;i++)
			//printf("%s\n",a[i]);
	    //printf("\n");	
		t=0;
		s++;
	}
	return 0;
}

//参考他人代码

单模式匹配算法：给定一个单词和一个字符串，查看字符串中是否存在该单词，通过调用strstr函数进行匹配；

#include<iostream>
#include<string>
#include<stdio.h>
#include<string.h>
 
using namespace std ;
 
int main()  
{
    char word[1000][20] ;
    int t = 1 ;
    while(cin >> word[0])  
	{
        int i = 1 ;
        while(cin >> word[i++])
            if(strcmp(word[i-1],"9")==0)
                break ;
        bool flag = true ;
        i--;
        for(int j = 0 ; j < i ; j++)//每一行均与其它进行比较 
		{
            char *p = NULL ;
            for(int k = 0 ; k < i ; k++)    
			{
                if(j == k)
                    continue ;
                p = strstr(word[j],word[k]) ;//第j行在其它行（k）中遍历，看是否是其前缀。
                 if(p == word[j]) 
	      {
                    flag = false ;
                    break ;
                }
            }
            if(flag == false)
                break ;
        }
        if(flag)
            printf("Set %d is immediately decodable\n",t++) ;
        else
            printf("Set %d is not immediately decodable\n",t++) ;
	}
    return 0 ;
}

转载于:https://www.cnblogs.com/NYNU-ACM/p/4237326.html