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  • Immediate Decodability

    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

    01
    10
    0010
    0000
    9
    01
    10
    010
    0000
    9

    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
    ",s);
    		else
                printf("Set %d is not immediately decodable
    ",s);
    	
    		//for(i=0;i<t;i++)
    			//printf("%s
    ",a[i]);
    	    //printf("
    ");	
    		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
    ",t++) ;
            else
                printf("Set %d is not immediately decodable
    ",t++) ;
    	}
        return 0 ;
    }
    
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  • 原文地址:https://www.cnblogs.com/NYNU-ACM/p/4237326.html
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