Immediate Decodability-字典树

Love The Way You Lie 2022-06-05 11:07 110阅读 0赞

# Immediate Decodability #

## Problem 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 input. 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.

## 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

题意：比较前缀，如果有一个字符串是另一个字符串的前缀，则这组数据is not immediately decodable。每组数据以‘9’分隔。

在建立字典树的过程中比较前缀，如果一个字符串结束后，它仍有儿子或已标记过。则有共同前缀。

[点击打开链接][Link 1]

#include<bits/stdc++.h>///Immediate Decodability
    using namespace std;
    const int maxn = 2;
    struct Node          ///字典树节点定义
    {
        int flag;
        Node *son[maxn];  ///2个儿子。
    };
    Node *newNode()
    {
        Node *p=new Node;
        p->flag=0;
        p->son[0]=NULL;
        p->son[1]=NULL;
        return p;
    }
    int gettree(Node *root,char *s)  ///将单词s插入到字典树
    {
        int i,l;
        l=strlen(s);
        Node *p = root;
        for(i=0;i<l;i++)
        {
            int k=s[i]-'0';
            if(p->flag==1)
            {
                return 0;
            }
            if(p->son[k]==NULL)
                p->son[k]=newNode();
            p=p->son[k];
        }
        if(p->son[0]==NULL&&p->son[1]==NULL&&p->flag==0)
        {
            p->flag=1;
            return 1;
        }
        else
            return 0;
    }
    void freedom(Node *p)
    {
        if(p->son[0])freedom(p->son[0]);
        if(p->son[1])freedom(p->son[1]);
        delete p;
    }
    int main()
    {
        int i,n=0,m=0;
        char s[20][20];
        Node *root;
        while(scanf("%s",s[n++])!=EOF)
        {
            if(s[n-1][0]!='9');
            else
            {
                n--;m++;
                root=newNode();
                for(i=0;i<n;i++)
                {
                    int k=gettree(root,s[i]);
                    if(k==0)break;
                }
                if(i==n)
                    printf("Set %d is immediately decodable\n", m);
                else
                    printf("Set %d is not immediately decodable\n", m);
                freedom(root);
                n = 0 ;
            }
        }
    }

[Link 1]: http://acm.hdu.edu.cn/webcontest/contest_showproblem.php?pid=1000&ojid=0&cid=12060&hide=0