寒假训练--字典树--A - Immediate Decodability

最新推荐文章于 2020-10-19 15:29:56 发布

原创最新推荐文章于 2020-10-19 15:29:56 发布 · 1.4k 阅读

0 ·

CC 4.0 BY-SA版权

本文介绍了一种用于检测一组符号编码是否立即可解的方法。通过构建二进制树并检查编码之间的前缀关系，来判断给定的编码集是否满足立即可解的条件。详细解释了输入格式、输出格式及算法实现过程。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Immediate Decodability

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

这个题就是判断不能有字符串是另一个字符串的前缀，首先将所有字符串存到数组中，在建树时判断有没有重复的，在循环中 if（p->next【k】）那么证明有一个字符串在这结束了，那个字符串就是正在循环的字符串的前缀，如果到最后，p->next 不为空，就证明这个字符串会是其他字符串的前缀

#include <stdio.h>
#include <string.h>
struct node
{
    int flag ;
    node *next[2] ;
};
node *newnode()
{
    node *p = new node ;
    p->flag = 0 ;
    p->next[0] = NULL ;
    p->next[1] = NULL ;
    return p ;
}
int gettree(node *root,char *s)
{
    int i , l = strlen(s);
    node *p = root ;
    for(i = 0 ; i < l ; i++)
    {
        int k = s[i]-'0' ;
        if(p->flag)
        {
            return 0;
        }
        if(p->next[k]==NULL)
            p->next[k] = newnode() ;
        p = p->next[k] ;
    }
    if(p->next[0] == NULL && p->next[1] == NULL && p->flag == 0)
    {
        p->flag = 1 ;
        return 1 ;
    }
    else
        return 0 ;
}
void freedom(node *p)
{
    if(p->next[0]) freedom(p->next[0]);
    if(p->next[1]) freedom(p->next[1]);
    delete p ;
}
int main()
{
    int i , n = 0 , num = 0;
    char s[20][20] ;
    node *head;
    while(scanf("%s", s[n++])!=EOF)
    {
        if(s[n-1][0] != '9') ;
        else
        {
            n-- ; num++;
            head = newnode();
            for(i = 0 ; i < n ; i++)
            {
                int k = gettree(head,s[i]);
                if(k==0) break;
            }
            if(i==n)
                printf("Set %d is immediately decodable\n", num);
            else
                printf("Set %d is not immediately decodable\n", num);
            freedom(head);
            n = 0 ;
        }
    }
}